An ice sled powered by a rocket engine sum from rest on a large frozen lake and accelerates at + 40 ft/s 2 . After some time t 1 , the rocket engine is shut down and the sled moves with constant velocity v for a time t 2 . If the total distance traveled by the sled is 17 500 ft and the total time is 90 s. find (a) the times t 1 and t 2 and (b) the velocity v . At the 17 500-ft mark, the sled begins to accelerate at −20 ft/s 2 . (c) What is the final position of the sled when it comes to rest? (d) How long does it take to come to rest?

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

Chapter 2, Problem 71AP

An ice sled powered by a rocket engine sum from rest on a large frozen lake and accelerates at + 40 ft/s². After some time t₁, the rocket engine is shut down and the sled moves with constant velocity v for a time t₂. If the total distance traveled by the sled is 17 500 ft and the total time is 90 s. find (a) the times t₁ and t₂ and (b) the velocity v. At the 17 500-ft mark, the sled begins to accelerate at −20 ft/s². (c) What is the final position of the sled when it comes to rest? (d) How long does it take to come to rest?

(a)

Expert Solution

To determine

The times

$t1$ and

$t2$ .

Answer to Problem 71AP

Solution: $t1$ and $t2$ are $5.0 s$ and $85.0 s$ respectively,

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the displacement of ice sled.

$Δx1=v0t1+12a1t12$

$Δx1$ is the distance travelled by ice sled with acceleration
$v0$ is the initial velocity of ice sled
$a1$ is the acceleration of ice sled
$t1$ is the time taken to cover the distance $Δx1$

Rewrite the above equation by substituting $0 ft/s$ for $v0$ and $40 ft/s2$ for $a$ .

$Δx1=(0 ft/s)t1+12(40 ft/s2)t12=(20 ft/s2)t12$ (I)

Write the formula to calculate the constant velocity of ice sled.

$v=v0+a1t1$

$v$ is the constant velocity gained by ice sled

Rewrite the above equation by substituting $0 ft/s$ for $v0$ and $40 ft/s2$ for $a$ .

$v=0 ft/s+(40 ft/s2)t1=(40 ft/s2)t1$ (II)

Write the formula to calculate the distance moved by ice sled with constant velocity.

$Δx2=vt2$

$Δx2$ is the distance moved by ice sled with constant velocity
$t2$ is the time taken to cover the distance $Δx2$

Rewrite the above equation by substituting equation (II).

$Δx2=(40 ft/s2)t1t2$ (III)

Write the relation between $Δx1$ and $Δx2$ .

$Δx1+Δx2=17500 ft$

Substitute equation (I) and equation (II) in the above relation.

$(20 ft/s2)t12+(40 ft/s2)t1t2=17500 ft$

Divide the above equation on both sides by $20 ft/s2$ .

$t12+2t1t2=875 s2$ (IV)

Write the relation to calculate the total time.

$t1+t2=90 s$

Rewrite the above equation in terms of $t2$ .

$t2=90 s−t1$ (V)

Rewrite equation (IV) by substituting equation (V).

$t12+2t1(90 s−t1)=875 s2t12−(180 s)t1+875 s2=0$

The above relation is a quadratic equation in variable $t1$ .

Solve the quadratic equation.

$t1=(180 s±(−180 s)2−4(1)(875 s2))2=(180 s±170 s2)=5.0 s or 175.0 s$

It is given that $t1+t2=90 s$ . This means $t1<90 s$ . So $t1=5.0 s$ is the required solution.

Substitute $5.0 s$ for $t1$ in equation (V) to find $t2$ .

$t2=90 s−5.0 s=85.0 s$

Conclusion: Thus, the times $t1$ and $t2$ are $5.0 s$ and $85.0 s$ respectively.

(b)

Expert Solution

To determine

The velocity

$v$ .

Answer to Problem 71AP

Solution: The velocity $v$ is $200.0 ft/s$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Substitute $5.0 s$ for $t1$ in equation (II) to find $v$ .

$v=(40 ft/s2)(5.0 s)=200.0 ft/s$

Conclusion: Thus, velocity $v$ is $200.0 ft/s$ .

(c)

Expert Solution

To determine

The total displacement of sled before it comes to rest.

Answer to Problem 71AP

Solution: The total displacement of sled is $18,500.0 ft$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the displacement of an object.

$Δx3=vf2−v22a2$

$Δx3$ is the displacement of sled when the acceleration is $−20 ft/s2$
$vf$ is the velocity of sled at the last point
$a2$ is the acceleration of sled before it comes to rest

Substitute $0 ft/s2$ for $vf$ , $200.0 ft/s$ for $v$ and $−20 ft/s2$ for $a2$ in the above equation to find $Δx3$ .

$Δx3=(0 ft/s2)2−(200.0 ft/s)22(−20 ft/s2)=1000.0 ft$

Write the formula to calculate the total displacement.

$Δxtotal=(Δx1+Δx2)+Δx3$

$Δxtotal$ is the total displacement of the ice sled

Substitute $17500 ft$ for $Δx1+Δx2$ and $1000.0 ft$ for $Δx3$ in the above equation to find $Δxtotal$ .

$Δxtotal=(17500 ft)+1000.0 ft=18,500.0 ft$

Conclusion: Thus, the total displacement of ice sled is $18,500.0 ft$ .

(d)

Expert Solution

To determine

The total time taken by ice sled to come to rest.

Answer to Problem 71AP

Solution: The time taken will be $10.0 s$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the time interval.

$t3=vf−va2$

$t3$ is the time taken by ice sled to come to rest from velocity $v$

Substitute $0 ft/s2$ for $vf$ , $200.0 ft/s$ for $v$ and the $−20 ft/s2$ for $a2$ in the above equation to find $t3$ .

$t3=0 ft/s2−200.0 ft/s−20 ft/s2=10.0 s$

Write the formula to calculate the total time duration of the trip.

$ttotal=t1+t2+t3$

$ttotal$ is the total time duration of the trip

Substitute $t1$ and $t2$ are $5.0 s$ for $t1$ , $85.0 s$ for $t2$ and $10.0 s$ for $t3$ in the above equation to find $ttotal$ .

$ttotal=5.0 s+85.0 s+10.0 s=100 s$

Conclusion: Thus, the ice sled comes to rest in $100 s$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

What is the error determined by the 2/3 rule?

Your colleague gives you a sample that are supposed to consist of Pt-Ni nanoparticles, TiO2 nanorod arrays, and SiO2 monolith plates (see right panel schematic). The bimetallic Pt-Ni nanoparticles are expected to decorate on the side surfaces of the aligned TiO2 nanorod arrays. These aligned TiO2 nanoarrays grew on the flat SiO2 monolith. Let's assume that the sizes of the Pt-Ni nanoparticles are > 10 nm. We further assume that you have access to a modern SEM that can produce a probe size as small as 1 nm with a current as high as 1 nA. You are not expected to damage/destroy the sample. Hint: keep your answers concise and to the point. TiO₂ Nanorods SiO, monolith a) What do you plan to do if your colleague wants to know if the Pt and Ni formed uniform alloy nanoparticles? (5 points) b) If your colleague wants to know the spatial distribution of the PtNi nanoparticles with respect to the TiO2 nanoarrays, how do you accomplish such a goal? (5 points) c) Based on the experimental results…

Find the current in 5.00 and 7.00 Ω resistors. Please explain all reasoning

Answer 2

Textbook Question

Chapter 2, Problem 71AP

An ice sled powered by a rocket engine sum from rest on a large frozen lake and accelerates at + 40 ft/s². After some time t₁, the rocket engine is shut down and the sled moves with constant velocity v for a time t₂. If the total distance traveled by the sled is 17 500 ft and the total time is 90 s. find (a) the times t₁ and t₂ and (b) the velocity v. At the 17 500-ft mark, the sled begins to accelerate at −20 ft/s². (c) What is the final position of the sled when it comes to rest? (d) How long does it take to come to rest?

(a)

Expert Solution

To determine

The times

$t1$ and

$t2$ .

Answer to Problem 71AP

Solution: $t1$ and $t2$ are $5.0 s$ and $85.0 s$ respectively,

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the displacement of ice sled.

$Δx1=v0t1+12a1t12$

$Δx1$ is the distance travelled by ice sled with acceleration
$v0$ is the initial velocity of ice sled
$a1$ is the acceleration of ice sled
$t1$ is the time taken to cover the distance $Δx1$

Rewrite the above equation by substituting $0 ft/s$ for $v0$ and $40 ft/s2$ for $a$ .

$Δx1=(0 ft/s)t1+12(40 ft/s2)t12=(20 ft/s2)t12$ (I)

Write the formula to calculate the constant velocity of ice sled.

$v=v0+a1t1$

$v$ is the constant velocity gained by ice sled

Rewrite the above equation by substituting $0 ft/s$ for $v0$ and $40 ft/s2$ for $a$ .

$v=0 ft/s+(40 ft/s2)t1=(40 ft/s2)t1$ (II)

Write the formula to calculate the distance moved by ice sled with constant velocity.

$Δx2=vt2$

$Δx2$ is the distance moved by ice sled with constant velocity
$t2$ is the time taken to cover the distance $Δx2$

Rewrite the above equation by substituting equation (II).

$Δx2=(40 ft/s2)t1t2$ (III)

Write the relation between $Δx1$ and $Δx2$ .

$Δx1+Δx2=17500 ft$

Substitute equation (I) and equation (II) in the above relation.

$(20 ft/s2)t12+(40 ft/s2)t1t2=17500 ft$

Divide the above equation on both sides by $20 ft/s2$ .

$t12+2t1t2=875 s2$ (IV)

Write the relation to calculate the total time.

$t1+t2=90 s$

Rewrite the above equation in terms of $t2$ .

$t2=90 s−t1$ (V)

Rewrite equation (IV) by substituting equation (V).

$t12+2t1(90 s−t1)=875 s2t12−(180 s)t1+875 s2=0$

The above relation is a quadratic equation in variable $t1$ .

Solve the quadratic equation.

$t1=(180 s±(−180 s)2−4(1)(875 s2))2=(180 s±170 s2)=5.0 s or 175.0 s$

It is given that $t1+t2=90 s$ . This means $t1<90 s$ . So $t1=5.0 s$ is the required solution.

Substitute $5.0 s$ for $t1$ in equation (V) to find $t2$ .

$t2=90 s−5.0 s=85.0 s$

Conclusion: Thus, the times $t1$ and $t2$ are $5.0 s$ and $85.0 s$ respectively.

(b)

Expert Solution

To determine

The velocity

$v$ .

Answer to Problem 71AP

Solution: The velocity $v$ is $200.0 ft/s$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Substitute $5.0 s$ for $t1$ in equation (II) to find $v$ .

$v=(40 ft/s2)(5.0 s)=200.0 ft/s$

Conclusion: Thus, velocity $v$ is $200.0 ft/s$ .

(c)

Expert Solution

To determine

The total displacement of sled before it comes to rest.

Answer to Problem 71AP

Solution: The total displacement of sled is $18,500.0 ft$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the displacement of an object.

$Δx3=vf2−v22a2$

$Δx3$ is the displacement of sled when the acceleration is $−20 ft/s2$
$vf$ is the velocity of sled at the last point
$a2$ is the acceleration of sled before it comes to rest

Substitute $0 ft/s2$ for $vf$ , $200.0 ft/s$ for $v$ and $−20 ft/s2$ for $a2$ in the above equation to find $Δx3$ .

$Δx3=(0 ft/s2)2−(200.0 ft/s)22(−20 ft/s2)=1000.0 ft$

Write the formula to calculate the total displacement.

$Δxtotal=(Δx1+Δx2)+Δx3$

$Δxtotal$ is the total displacement of the ice sled

Substitute $17500 ft$ for $Δx1+Δx2$ and $1000.0 ft$ for $Δx3$ in the above equation to find $Δxtotal$ .

$Δxtotal=(17500 ft)+1000.0 ft=18,500.0 ft$

Conclusion: Thus, the total displacement of ice sled is $18,500.0 ft$ .

(d)

Expert Solution

To determine

The total time taken by ice sled to come to rest.

Answer to Problem 71AP

Solution: The time taken will be $10.0 s$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the time interval.

$t3=vf−va2$

$t3$ is the time taken by ice sled to come to rest from velocity $v$

Substitute $0 ft/s2$ for $vf$ , $200.0 ft/s$ for $v$ and the $−20 ft/s2$ for $a2$ in the above equation to find $t3$ .

$t3=0 ft/s2−200.0 ft/s−20 ft/s2=10.0 s$

Write the formula to calculate the total time duration of the trip.

$ttotal=t1+t2+t3$

$ttotal$ is the total time duration of the trip

Substitute $t1$ and $t2$ are $5.0 s$ for $t1$ , $85.0 s$ for $t2$ and $10.0 s$ for $t3$ in the above equation to find $ttotal$ .

$ttotal=5.0 s+85.0 s+10.0 s=100 s$

Conclusion: Thus, the ice sled comes to rest in $100 s$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

Chapter 2, Problem 71AP

An ice sled powered by a rocket engine sum from rest on a large frozen lake and accelerates at + 40 ft/s². After some time t₁, the rocket engine is shut down and the sled moves with constant velocity v for a time t₂. If the total distance traveled by the sled is 17 500 ft and the total time is 90 s. find (a) the times t₁ and t₂ and (b) the velocity v. At the 17 500-ft mark, the sled begins to accelerate at −20 ft/s². (c) What is the final position of the sled when it comes to rest? (d) How long does it take to come to rest?

(a)

Expert Solution

To determine

The times

$t1$ and

$t2$ .

Answer to Problem 71AP

Solution: $t1$ and $t2$ are $5.0 s$ and $85.0 s$ respectively,

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the displacement of ice sled.

$Δx1=v0t1+12a1t12$

$Δx1$ is the distance travelled by ice sled with acceleration
$v0$ is the initial velocity of ice sled
$a1$ is the acceleration of ice sled
$t1$ is the time taken to cover the distance $Δx1$

Rewrite the above equation by substituting $0 ft/s$ for $v0$ and $40 ft/s2$ for $a$ .

$Δx1=(0 ft/s)t1+12(40 ft/s2)t12=(20 ft/s2)t12$ (I)

Write the formula to calculate the constant velocity of ice sled.

$v=v0+a1t1$

$v$ is the constant velocity gained by ice sled

Rewrite the above equation by substituting $0 ft/s$ for $v0$ and $40 ft/s2$ for $a$ .

$v=0 ft/s+(40 ft/s2)t1=(40 ft/s2)t1$ (II)

Write the formula to calculate the distance moved by ice sled with constant velocity.

$Δx2=vt2$

$Δx2$ is the distance moved by ice sled with constant velocity
$t2$ is the time taken to cover the distance $Δx2$

Rewrite the above equation by substituting equation (II).

$Δx2=(40 ft/s2)t1t2$ (III)

Write the relation between $Δx1$ and $Δx2$ .

$Δx1+Δx2=17500 ft$

Substitute equation (I) and equation (II) in the above relation.

$(20 ft/s2)t12+(40 ft/s2)t1t2=17500 ft$

Divide the above equation on both sides by $20 ft/s2$ .

$t12+2t1t2=875 s2$ (IV)

Write the relation to calculate the total time.

$t1+t2=90 s$

Rewrite the above equation in terms of $t2$ .

$t2=90 s−t1$ (V)

Rewrite equation (IV) by substituting equation (V).

$t12+2t1(90 s−t1)=875 s2t12−(180 s)t1+875 s2=0$

The above relation is a quadratic equation in variable $t1$ .

Solve the quadratic equation.

$t1=(180 s±(−180 s)2−4(1)(875 s2))2=(180 s±170 s2)=5.0 s or 175.0 s$

It is given that $t1+t2=90 s$ . This means $t1<90 s$ . So $t1=5.0 s$ is the required solution.

Substitute $5.0 s$ for $t1$ in equation (V) to find $t2$ .

$t2=90 s−5.0 s=85.0 s$

Conclusion: Thus, the times $t1$ and $t2$ are $5.0 s$ and $85.0 s$ respectively.

(b)

Expert Solution

To determine

The velocity

$v$ .

Answer to Problem 71AP

Solution: The velocity $v$ is $200.0 ft/s$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Substitute $5.0 s$ for $t1$ in equation (II) to find $v$ .

$v=(40 ft/s2)(5.0 s)=200.0 ft/s$

Conclusion: Thus, velocity $v$ is $200.0 ft/s$ .

(c)

Expert Solution

To determine

The total displacement of sled before it comes to rest.

Answer to Problem 71AP

Solution: The total displacement of sled is $18,500.0 ft$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the displacement of an object.

$Δx3=vf2−v22a2$

$Δx3$ is the displacement of sled when the acceleration is $−20 ft/s2$
$vf$ is the velocity of sled at the last point
$a2$ is the acceleration of sled before it comes to rest

Substitute $0 ft/s2$ for $vf$ , $200.0 ft/s$ for $v$ and $−20 ft/s2$ for $a2$ in the above equation to find $Δx3$ .

$Δx3=(0 ft/s2)2−(200.0 ft/s)22(−20 ft/s2)=1000.0 ft$

Write the formula to calculate the total displacement.

$Δxtotal=(Δx1+Δx2)+Δx3$

$Δxtotal$ is the total displacement of the ice sled

Substitute $17500 ft$ for $Δx1+Δx2$ and $1000.0 ft$ for $Δx3$ in the above equation to find $Δxtotal$ .

$Δxtotal=(17500 ft)+1000.0 ft=18,500.0 ft$

Conclusion: Thus, the total displacement of ice sled is $18,500.0 ft$ .

(d)

Expert Solution

To determine

The total time taken by ice sled to come to rest.

Answer to Problem 71AP

Solution: The time taken will be $10.0 s$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the time interval.

$t3=vf−va2$

$t3$ is the time taken by ice sled to come to rest from velocity $v$

Substitute $0 ft/s2$ for $vf$ , $200.0 ft/s$ for $v$ and the $−20 ft/s2$ for $a2$ in the above equation to find $t3$ .

$t3=0 ft/s2−200.0 ft/s−20 ft/s2=10.0 s$

Write the formula to calculate the total time duration of the trip.

$ttotal=t1+t2+t3$

$ttotal$ is the total time duration of the trip

Substitute $t1$ and $t2$ are $5.0 s$ for $t1$ , $85.0 s$ for $t2$ and $10.0 s$ for $t3$ in the above equation to find $ttotal$ .

$ttotal=5.0 s+85.0 s+10.0 s=100 s$

Conclusion: Thus, the ice sled comes to rest in $100 s$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 2, Problem 71AP

An ice sled powered by a rocket engine sum from rest on a large frozen lake and accelerates at + 40 ft/s². After some time t₁, the rocket engine is shut down and the sled moves with constant velocity v for a time t₂. If the total distance traveled by the sled is 17 500 ft and the total time is 90 s. find (a) the times t₁ and t₂ and (b) the velocity v. At the 17 500-ft mark, the sled begins to accelerate at −20 ft/s². (c) What is the final position of the sled when it comes to rest? (d) How long does it take to come to rest?

(a)

Expert Solution

To determine

The times

$t1$ and

$t2$ .

Answer to Problem 71AP

Solution: $t1$ and $t2$ are $5.0 s$ and $85.0 s$ respectively,

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the displacement of ice sled.

$Δx1=v0t1+12a1t12$

$Δx1$ is the distance travelled by ice sled with acceleration
$v0$ is the initial velocity of ice sled
$a1$ is the acceleration of ice sled
$t1$ is the time taken to cover the distance $Δx1$

Rewrite the above equation by substituting $0 ft/s$ for $v0$ and $40 ft/s2$ for $a$ .

$Δx1=(0 ft/s)t1+12(40 ft/s2)t12=(20 ft/s2)t12$ (I)

Write the formula to calculate the constant velocity of ice sled.

$v=v0+a1t1$

$v$ is the constant velocity gained by ice sled

Rewrite the above equation by substituting $0 ft/s$ for $v0$ and $40 ft/s2$ for $a$ .

$v=0 ft/s+(40 ft/s2)t1=(40 ft/s2)t1$ (II)

Write the formula to calculate the distance moved by ice sled with constant velocity.

$Δx2=vt2$

$Δx2$ is the distance moved by ice sled with constant velocity
$t2$ is the time taken to cover the distance $Δx2$

Rewrite the above equation by substituting equation (II).

$Δx2=(40 ft/s2)t1t2$ (III)

Write the relation between $Δx1$ and $Δx2$ .

$Δx1+Δx2=17500 ft$

Substitute equation (I) and equation (II) in the above relation.

$(20 ft/s2)t12+(40 ft/s2)t1t2=17500 ft$

Divide the above equation on both sides by $20 ft/s2$ .

$t12+2t1t2=875 s2$ (IV)

Write the relation to calculate the total time.

$t1+t2=90 s$

Rewrite the above equation in terms of $t2$ .

$t2=90 s−t1$ (V)

Rewrite equation (IV) by substituting equation (V).

$t12+2t1(90 s−t1)=875 s2t12−(180 s)t1+875 s2=0$

The above relation is a quadratic equation in variable $t1$ .

Solve the quadratic equation.

$t1=(180 s±(−180 s)2−4(1)(875 s2))2=(180 s±170 s2)=5.0 s or 175.0 s$

It is given that $t1+t2=90 s$ . This means $t1<90 s$ . So $t1=5.0 s$ is the required solution.

Substitute $5.0 s$ for $t1$ in equation (V) to find $t2$ .

$t2=90 s−5.0 s=85.0 s$

Conclusion: Thus, the times $t1$ and $t2$ are $5.0 s$ and $85.0 s$ respectively.

(b)

Expert Solution

To determine

The velocity

$v$ .

Answer to Problem 71AP

Solution: The velocity $v$ is $200.0 ft/s$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Substitute $5.0 s$ for $t1$ in equation (II) to find $v$ .

$v=(40 ft/s2)(5.0 s)=200.0 ft/s$

Conclusion: Thus, velocity $v$ is $200.0 ft/s$ .

(c)

Expert Solution

To determine

The total displacement of sled before it comes to rest.

Answer to Problem 71AP

Solution: The total displacement of sled is $18,500.0 ft$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the displacement of an object.

$Δx3=vf2−v22a2$

$Δx3$ is the displacement of sled when the acceleration is $−20 ft/s2$
$vf$ is the velocity of sled at the last point
$a2$ is the acceleration of sled before it comes to rest

Substitute $0 ft/s2$ for $vf$ , $200.0 ft/s$ for $v$ and $−20 ft/s2$ for $a2$ in the above equation to find $Δx3$ .

$Δx3=(0 ft/s2)2−(200.0 ft/s)22(−20 ft/s2)=1000.0 ft$

Write the formula to calculate the total displacement.

$Δxtotal=(Δx1+Δx2)+Δx3$

$Δxtotal$ is the total displacement of the ice sled

Substitute $17500 ft$ for $Δx1+Δx2$ and $1000.0 ft$ for $Δx3$ in the above equation to find $Δxtotal$ .

$Δxtotal=(17500 ft)+1000.0 ft=18,500.0 ft$

Conclusion: Thus, the total displacement of ice sled is $18,500.0 ft$ .

(d)

Expert Solution

To determine

The total time taken by ice sled to come to rest.

Answer to Problem 71AP

Solution: The time taken will be $10.0 s$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the time interval.

$t3=vf−va2$

$t3$ is the time taken by ice sled to come to rest from velocity $v$

Substitute $0 ft/s2$ for $vf$ , $200.0 ft/s$ for $v$ and the $−20 ft/s2$ for $a2$ in the above equation to find $t3$ .

$t3=0 ft/s2−200.0 ft/s−20 ft/s2=10.0 s$

Write the formula to calculate the total time duration of the trip.

$ttotal=t1+t2+t3$

$ttotal$ is the total time duration of the trip

Substitute $t1$ and $t2$ are $5.0 s$ for $t1$ , $85.0 s$ for $t2$ and $10.0 s$ for $t3$ in the above equation to find $ttotal$ .

$ttotal=5.0 s+85.0 s+10.0 s=100 s$

Conclusion: Thus, the ice sled comes to rest in $100 s$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The times

$t1$ and

$t2$ .

Answer to Problem 71AP

Solution: $t1$ and $t2$ are $5.0 s$ and $85.0 s$ respectively,

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the displacement of ice sled.

$Δx1=v0t1+12a1t12$

$Δx1$ is the distance travelled by ice sled with acceleration
$v0$ is the initial velocity of ice sled
$a1$ is the acceleration of ice sled
$t1$ is the time taken to cover the distance $Δx1$

Rewrite the above equation by substituting $0 ft/s$ for $v0$ and $40 ft/s2$ for $a$ .

$Δx1=(0 ft/s)t1+12(40 ft/s2)t12=(20 ft/s2)t12$ (I)

Write the formula to calculate the constant velocity of ice sled.

$v=v0+a1t1$

$v$ is the constant velocity gained by ice sled

Rewrite the above equation by substituting $0 ft/s$ for $v0$ and $40 ft/s2$ for $a$ .

$v=0 ft/s+(40 ft/s2)t1=(40 ft/s2)t1$ (II)

Write the formula to calculate the distance moved by ice sled with constant velocity.

$Δx2=vt2$

$Δx2$ is the distance moved by ice sled with constant velocity
$t2$ is the time taken to cover the distance $Δx2$

Rewrite the above equation by substituting equation (II).

$Δx2=(40 ft/s2)t1t2$ (III)

Write the relation between $Δx1$ and $Δx2$ .

$Δx1+Δx2=17500 ft$

Substitute equation (I) and equation (II) in the above relation.

$(20 ft/s2)t12+(40 ft/s2)t1t2=17500 ft$

Divide the above equation on both sides by $20 ft/s2$ .

$t12+2t1t2=875 s2$ (IV)

Write the relation to calculate the total time.

$t1+t2=90 s$

Rewrite the above equation in terms of $t2$ .

$t2=90 s−t1$ (V)

Rewrite equation (IV) by substituting equation (V).

$t12+2t1(90 s−t1)=875 s2t12−(180 s)t1+875 s2=0$

The above relation is a quadratic equation in variable $t1$ .

Solve the quadratic equation.

$t1=(180 s±(−180 s)2−4(1)(875 s2))2=(180 s±170 s2)=5.0 s or 175.0 s$

It is given that $t1+t2=90 s$ . This means $t1<90 s$ . So $t1=5.0 s$ is the required solution.

Substitute $5.0 s$ for $t1$ in equation (V) to find $t2$ .

$t2=90 s−5.0 s=85.0 s$

Conclusion: Thus, the times $t1$ and $t2$ are $5.0 s$ and $85.0 s$ respectively.

(b)

Expert Solution

To determine

The velocity

$v$ .

Answer to Problem 71AP

Solution: The velocity $v$ is $200.0 ft/s$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Substitute $5.0 s$ for $t1$ in equation (II) to find $v$ .

$v=(40 ft/s2)(5.0 s)=200.0 ft/s$

Conclusion: Thus, velocity $v$ is $200.0 ft/s$ .

(c)

Expert Solution

To determine

The total displacement of sled before it comes to rest.

Answer to Problem 71AP

Solution: The total displacement of sled is $18,500.0 ft$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the displacement of an object.

$Δx3=vf2−v22a2$

$Δx3$ is the displacement of sled when the acceleration is $−20 ft/s2$
$vf$ is the velocity of sled at the last point
$a2$ is the acceleration of sled before it comes to rest

Substitute $0 ft/s2$ for $vf$ , $200.0 ft/s$ for $v$ and $−20 ft/s2$ for $a2$ in the above equation to find $Δx3$ .

$Δx3=(0 ft/s2)2−(200.0 ft/s)22(−20 ft/s2)=1000.0 ft$

Write the formula to calculate the total displacement.

$Δxtotal=(Δx1+Δx2)+Δx3$

$Δxtotal$ is the total displacement of the ice sled

Substitute $17500 ft$ for $Δx1+Δx2$ and $1000.0 ft$ for $Δx3$ in the above equation to find $Δxtotal$ .

$Δxtotal=(17500 ft)+1000.0 ft=18,500.0 ft$

Conclusion: Thus, the total displacement of ice sled is $18,500.0 ft$ .

(d)

Expert Solution

To determine

The total time taken by ice sled to come to rest.

Answer to Problem 71AP

Solution: The time taken will be $10.0 s$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the time interval.

$t3=vf−va2$

$t3$ is the time taken by ice sled to come to rest from velocity $v$

Substitute $0 ft/s2$ for $vf$ , $200.0 ft/s$ for $v$ and the $−20 ft/s2$ for $a2$ in the above equation to find $t3$ .

$t3=0 ft/s2−200.0 ft/s−20 ft/s2=10.0 s$

Write the formula to calculate the total time duration of the trip.

$ttotal=t1+t2+t3$

$ttotal$ is the total time duration of the trip

Substitute $t1$ and $t2$ are $5.0 s$ for $t1$ , $85.0 s$ for $t2$ and $10.0 s$ for $t3$ in the above equation to find $ttotal$ .

$ttotal=5.0 s+85.0 s+10.0 s=100 s$

Conclusion: Thus, the ice sled comes to rest in $100 s$

Answer 8

(a)

Expert Solution

To determine

The times

$t1$ and

$t2$ .

Answer to Problem 71AP

Solution: $t1$ and $t2$ are $5.0 s$ and $85.0 s$ respectively,

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the displacement of ice sled.

$Δx1=v0t1+12a1t12$

$Δx1$ is the distance travelled by ice sled with acceleration
$v0$ is the initial velocity of ice sled
$a1$ is the acceleration of ice sled
$t1$ is the time taken to cover the distance $Δx1$

Rewrite the above equation by substituting $0 ft/s$ for $v0$ and $40 ft/s2$ for $a$ .

$Δx1=(0 ft/s)t1+12(40 ft/s2)t12=(20 ft/s2)t12$ (I)

Write the formula to calculate the constant velocity of ice sled.

$v=v0+a1t1$

$v$ is the constant velocity gained by ice sled

Rewrite the above equation by substituting $0 ft/s$ for $v0$ and $40 ft/s2$ for $a$ .

$v=0 ft/s+(40 ft/s2)t1=(40 ft/s2)t1$ (II)

Write the formula to calculate the distance moved by ice sled with constant velocity.

$Δx2=vt2$

$Δx2$ is the distance moved by ice sled with constant velocity
$t2$ is the time taken to cover the distance $Δx2$

Rewrite the above equation by substituting equation (II).

$Δx2=(40 ft/s2)t1t2$ (III)

Write the relation between $Δx1$ and $Δx2$ .

$Δx1+Δx2=17500 ft$

Substitute equation (I) and equation (II) in the above relation.

$(20 ft/s2)t12+(40 ft/s2)t1t2=17500 ft$

Divide the above equation on both sides by $20 ft/s2$ .

$t12+2t1t2=875 s2$ (IV)

Write the relation to calculate the total time.

$t1+t2=90 s$

Rewrite the above equation in terms of $t2$ .

$t2=90 s−t1$ (V)

Rewrite equation (IV) by substituting equation (V).

$t12+2t1(90 s−t1)=875 s2t12−(180 s)t1+875 s2=0$

The above relation is a quadratic equation in variable $t1$ .

Solve the quadratic equation.

$t1=(180 s±(−180 s)2−4(1)(875 s2))2=(180 s±170 s2)=5.0 s or 175.0 s$

It is given that $t1+t2=90 s$ . This means $t1<90 s$ . So $t1=5.0 s$ is the required solution.

Substitute $5.0 s$ for $t1$ in equation (V) to find $t2$ .

$t2=90 s−5.0 s=85.0 s$

Conclusion: Thus, the times $t1$ and $t2$ are $5.0 s$ and $85.0 s$ respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The times

$t1$ and

$t2$ .

Answer 13

Answer to Problem 71AP

Solution: $t1$ and $t2$ are $5.0 s$ and $85.0 s$ respectively,

Answer 14

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the displacement of ice sled.

$Δx1=v0t1+12a1t12$

$Δx1$ is the distance travelled by ice sled with acceleration
$v0$ is the initial velocity of ice sled
$a1$ is the acceleration of ice sled
$t1$ is the time taken to cover the distance $Δx1$

Rewrite the above equation by substituting $0 ft/s$ for $v0$ and $40 ft/s2$ for $a$ .

$Δx1=(0 ft/s)t1+12(40 ft/s2)t12=(20 ft/s2)t12$ (I)

Write the formula to calculate the constant velocity of ice sled.

$v=v0+a1t1$

$v$ is the constant velocity gained by ice sled

Rewrite the above equation by substituting $0 ft/s$ for $v0$ and $40 ft/s2$ for $a$ .

$v=0 ft/s+(40 ft/s2)t1=(40 ft/s2)t1$ (II)

Write the formula to calculate the distance moved by ice sled with constant velocity.

$Δx2=vt2$

$Δx2$ is the distance moved by ice sled with constant velocity
$t2$ is the time taken to cover the distance $Δx2$

Rewrite the above equation by substituting equation (II).

$Δx2=(40 ft/s2)t1t2$ (III)

Write the relation between $Δx1$ and $Δx2$ .

$Δx1+Δx2=17500 ft$

Substitute equation (I) and equation (II) in the above relation.

$(20 ft/s2)t12+(40 ft/s2)t1t2=17500 ft$

Divide the above equation on both sides by $20 ft/s2$ .

$t12+2t1t2=875 s2$ (IV)

Write the relation to calculate the total time.

$t1+t2=90 s$

Rewrite the above equation in terms of $t2$ .

$t2=90 s−t1$ (V)

Rewrite equation (IV) by substituting equation (V).

$t12+2t1(90 s−t1)=875 s2t12−(180 s)t1+875 s2=0$

The above relation is a quadratic equation in variable $t1$ .

Solve the quadratic equation.

$t1=(180 s±(−180 s)2−4(1)(875 s2))2=(180 s±170 s2)=5.0 s or 175.0 s$

It is given that $t1+t2=90 s$ . This means $t1<90 s$ . So $t1=5.0 s$ is the required solution.

Substitute $5.0 s$ for $t1$ in equation (V) to find $t2$ .

$t2=90 s−5.0 s=85.0 s$

Conclusion: Thus, the times $t1$ and $t2$ are $5.0 s$ and $85.0 s$ respectively.

Answer 15

(b)

Expert Solution

To determine

The velocity

$v$ .

Answer to Problem 71AP

Solution: The velocity $v$ is $200.0 ft/s$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Substitute $5.0 s$ for $t1$ in equation (II) to find $v$ .

$v=(40 ft/s2)(5.0 s)=200.0 ft/s$

Conclusion: Thus, velocity $v$ is $200.0 ft/s$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The velocity

$v$ .

Answer 20

Answer to Problem 71AP

Solution: The velocity $v$ is $200.0 ft/s$ .

Answer 21

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Substitute $5.0 s$ for $t1$ in equation (II) to find $v$ .

$v=(40 ft/s2)(5.0 s)=200.0 ft/s$

Conclusion: Thus, velocity $v$ is $200.0 ft/s$ .

Answer 22

(c)

Expert Solution

To determine

The total displacement of sled before it comes to rest.

Answer to Problem 71AP

Solution: The total displacement of sled is $18,500.0 ft$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the displacement of an object.

$Δx3=vf2−v22a2$

$Δx3$ is the displacement of sled when the acceleration is $−20 ft/s2$
$vf$ is the velocity of sled at the last point
$a2$ is the acceleration of sled before it comes to rest

Substitute $0 ft/s2$ for $vf$ , $200.0 ft/s$ for $v$ and $−20 ft/s2$ for $a2$ in the above equation to find $Δx3$ .

$Δx3=(0 ft/s2)2−(200.0 ft/s)22(−20 ft/s2)=1000.0 ft$

Write the formula to calculate the total displacement.

$Δxtotal=(Δx1+Δx2)+Δx3$

$Δxtotal$ is the total displacement of the ice sled

Substitute $17500 ft$ for $Δx1+Δx2$ and $1000.0 ft$ for $Δx3$ in the above equation to find $Δxtotal$ .

$Δxtotal=(17500 ft)+1000.0 ft=18,500.0 ft$

Conclusion: Thus, the total displacement of ice sled is $18,500.0 ft$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The total displacement of sled before it comes to rest.

Answer 27

Answer to Problem 71AP

Solution: The total displacement of sled is $18,500.0 ft$ .

Answer 28

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the displacement of an object.

$Δx3=vf2−v22a2$

$Δx3$ is the displacement of sled when the acceleration is $−20 ft/s2$
$vf$ is the velocity of sled at the last point
$a2$ is the acceleration of sled before it comes to rest

Substitute $0 ft/s2$ for $vf$ , $200.0 ft/s$ for $v$ and $−20 ft/s2$ for $a2$ in the above equation to find $Δx3$ .

$Δx3=(0 ft/s2)2−(200.0 ft/s)22(−20 ft/s2)=1000.0 ft$

Write the formula to calculate the total displacement.

$Δxtotal=(Δx1+Δx2)+Δx3$

$Δxtotal$ is the total displacement of the ice sled

Substitute $17500 ft$ for $Δx1+Δx2$ and $1000.0 ft$ for $Δx3$ in the above equation to find $Δxtotal$ .

$Δxtotal=(17500 ft)+1000.0 ft=18,500.0 ft$

Conclusion: Thus, the total displacement of ice sled is $18,500.0 ft$ .

Answer 29

(d)

Expert Solution

To determine

The total time taken by ice sled to come to rest.

Answer to Problem 71AP

Solution: The time taken will be $10.0 s$ .

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .

Write the formula to calculate the time interval.

$t3=vf−va2$

$t3$ is the time taken by ice sled to come to rest from velocity $v$

Substitute $0 ft/s2$ for $vf$ , $200.0 ft/s$ for $v$ and the $−20 ft/s2$ for $a2$ in the above equation to find $t3$ .

$t3=0 ft/s2−200.0 ft/s−20 ft/s2=10.0 s$

Write the formula to calculate the total time duration of the trip.

$ttotal=t1+t2+t3$

$ttotal$ is the total time duration of the trip

Substitute $t1$ and $t2$ are $5.0 s$ for $t1$ , $85.0 s$ for $t2$ and $10.0 s$ for $t3$ in the above equation to find $ttotal$ .

$ttotal=5.0 s+85.0 s+10.0 s=100 s$

Conclusion: Thus, the ice sled comes to rest in $100 s$

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The total time taken by ice sled to come to rest.

Answer 34

Answer to Problem 71AP

Solution: The time taken will be $10.0 s$ .

Answer 35

Explanation of Solution

Given Info: The acceleration is $40 ft/s2$ , total distance is $17500 ft$ , the total time is $90 s$ and the acceleration after $17500 ft$ is $−20 ft/s2$ .