The velocity of the car after each second.

Question

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Answer 1

Question

Chapter 2, Problem 18E

(a)

To determine

The velocity of the car after each second.

(a)

Expert Solution

Answer to Problem 18E

The velocities after 1s, 2s, 3s, 4s and 5s are $3.0 m/s$ , $6.0 m/s$ , $9.0 m/s$ , $12 m/s$ and $15 m/s$ respectively.

Explanation of Solution

Given info: The car accelerates at $3.0 m/s2$ and the time is 5 seconds.

Write the formula for average acceleration.

$a=vf−vit$

Here,

a is the average acceleration

$vf$ is the final velocity

$vi$ is the initial velocity

t is the time

Re-arrange the equation to get $vf$ .

$vf=vi+at$

For $t=1 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and 1 s for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(1 s)=3.0 m/s$

For $t=2 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $2 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(2 s)=6.0 m/s$

For $t=3 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $3 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(3 s)=9.0 m/s$

For $t=4 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $4 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(4 s)=12 m/s$

For $t=5 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $5 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(5 s)=15 m/s$

The velocity time graph is given below.

The Physics of Everyday Phenomena, Chapter 2, Problem 18E , additional homework tip 1

Figure (1)

Conclusion:

Therefore, the velocities after 1s, 2s, 3s, 4s and 5s are $3.0 m/s$ , $6.0 m/s$ , $9.0 m/s$ , $12 m/s$ and $15 m/s$ respectively.

(b)

To determine

The distance travelled after each second.

(b)

Expert Solution

Answer to Problem 18E

The distances after 1s, 2s, 3s, 4s and 5s are $1.5 m$ , $6.0 m$ , $13.5 m$ , $24 m$ and $37.5 m$ respectively.

Explanation of Solution

Given info: Acceleration is $4 m/s2$ and time is 5 seconds.

Write the formula to calculate the distance.

$d=v0t+12at2$

Here,

d is the distance

For $t=1 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $1 s$ for t to get $d$ .

$d=(0 m/s)(1 s)+12(3.0 m/s2)(1 s)2=1.5 m$

For $t=2 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $2 s$ for t to get $d$ .

$d=(0 m/s)(2 s)+12(3.0 m/s2)(2 s)2=6.0 m$

For $t=3 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $3 s$ for t to get $d$ .

$d=(0 m/s)(3 s)+12(3.0 m/s2)(3 s)2=13.5 m$

For $t=4 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $4 s$ for t to get $d$ .

$d=(0 m/s)(4 s)+12(3.0 m/s2)(4 s)2=24 m$

For $t=5 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $5 s$ for t to get $d$ .

$d=(0 m/s)(5 s)+12(3.0 m/s2)(5 s)2=37.5 m$

The distance time graph is,

The Physics of Everyday Phenomena, Chapter 2, Problem 18E , additional homework tip 2

Figure(2)

Conclusion:

Therefore, the distances after 1s, 2s, 3s, 4s and 5s are $1.5 m$ , $6.0 m$ , $13.5 m$ , $24 m$ and $37.5 m$ respectively.

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Answer 2

Question

Chapter 2, Problem 18E

(a)

To determine

The velocity of the car after each second.

(a)

Expert Solution

Answer to Problem 18E

The velocities after 1s, 2s, 3s, 4s and 5s are $3.0 m/s$ , $6.0 m/s$ , $9.0 m/s$ , $12 m/s$ and $15 m/s$ respectively.

Explanation of Solution

Given info: The car accelerates at $3.0 m/s2$ and the time is 5 seconds.

Write the formula for average acceleration.

$a=vf−vit$

Here,

a is the average acceleration

$vf$ is the final velocity

$vi$ is the initial velocity

t is the time

Re-arrange the equation to get $vf$ .

$vf=vi+at$

For $t=1 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and 1 s for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(1 s)=3.0 m/s$

For $t=2 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $2 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(2 s)=6.0 m/s$

For $t=3 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $3 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(3 s)=9.0 m/s$

For $t=4 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $4 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(4 s)=12 m/s$

For $t=5 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $5 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(5 s)=15 m/s$

The velocity time graph is given below.

The Physics of Everyday Phenomena, Chapter 2, Problem 18E , additional homework tip 1

Figure (1)

Conclusion:

Therefore, the velocities after 1s, 2s, 3s, 4s and 5s are $3.0 m/s$ , $6.0 m/s$ , $9.0 m/s$ , $12 m/s$ and $15 m/s$ respectively.

(b)

To determine

The distance travelled after each second.

(b)

Expert Solution

Answer to Problem 18E

The distances after 1s, 2s, 3s, 4s and 5s are $1.5 m$ , $6.0 m$ , $13.5 m$ , $24 m$ and $37.5 m$ respectively.

Explanation of Solution

Given info: Acceleration is $4 m/s2$ and time is 5 seconds.

Write the formula to calculate the distance.

$d=v0t+12at2$

Here,

d is the distance

For $t=1 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $1 s$ for t to get $d$ .

$d=(0 m/s)(1 s)+12(3.0 m/s2)(1 s)2=1.5 m$

For $t=2 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $2 s$ for t to get $d$ .

$d=(0 m/s)(2 s)+12(3.0 m/s2)(2 s)2=6.0 m$

For $t=3 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $3 s$ for t to get $d$ .

$d=(0 m/s)(3 s)+12(3.0 m/s2)(3 s)2=13.5 m$

For $t=4 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $4 s$ for t to get $d$ .

$d=(0 m/s)(4 s)+12(3.0 m/s2)(4 s)2=24 m$

For $t=5 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $5 s$ for t to get $d$ .

$d=(0 m/s)(5 s)+12(3.0 m/s2)(5 s)2=37.5 m$

The distance time graph is,

The Physics of Everyday Phenomena, Chapter 2, Problem 18E , additional homework tip 2

Figure(2)

Conclusion:

Therefore, the distances after 1s, 2s, 3s, 4s and 5s are $1.5 m$ , $6.0 m$ , $13.5 m$ , $24 m$ and $37.5 m$ respectively.

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Answer 3

Question

Chapter 2, Problem 18E

(a)

To determine

The velocity of the car after each second.

(a)

Expert Solution

Answer to Problem 18E

The velocities after 1s, 2s, 3s, 4s and 5s are $3.0 m/s$ , $6.0 m/s$ , $9.0 m/s$ , $12 m/s$ and $15 m/s$ respectively.

Explanation of Solution

Given info: The car accelerates at $3.0 m/s2$ and the time is 5 seconds.

Write the formula for average acceleration.

$a=vf−vit$

Here,

a is the average acceleration

$vf$ is the final velocity

$vi$ is the initial velocity

t is the time

Re-arrange the equation to get $vf$ .

$vf=vi+at$

For $t=1 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and 1 s for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(1 s)=3.0 m/s$

For $t=2 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $2 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(2 s)=6.0 m/s$

For $t=3 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $3 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(3 s)=9.0 m/s$

For $t=4 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $4 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(4 s)=12 m/s$

For $t=5 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $5 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(5 s)=15 m/s$

The velocity time graph is given below.

The Physics of Everyday Phenomena, Chapter 2, Problem 18E , additional homework tip 1

Figure (1)

Conclusion:

Therefore, the velocities after 1s, 2s, 3s, 4s and 5s are $3.0 m/s$ , $6.0 m/s$ , $9.0 m/s$ , $12 m/s$ and $15 m/s$ respectively.

(b)

To determine

The distance travelled after each second.

(b)

Expert Solution

Answer to Problem 18E

The distances after 1s, 2s, 3s, 4s and 5s are $1.5 m$ , $6.0 m$ , $13.5 m$ , $24 m$ and $37.5 m$ respectively.

Explanation of Solution

Given info: Acceleration is $4 m/s2$ and time is 5 seconds.

Write the formula to calculate the distance.

$d=v0t+12at2$

Here,

d is the distance

For $t=1 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $1 s$ for t to get $d$ .

$d=(0 m/s)(1 s)+12(3.0 m/s2)(1 s)2=1.5 m$

For $t=2 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $2 s$ for t to get $d$ .

$d=(0 m/s)(2 s)+12(3.0 m/s2)(2 s)2=6.0 m$

For $t=3 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $3 s$ for t to get $d$ .

$d=(0 m/s)(3 s)+12(3.0 m/s2)(3 s)2=13.5 m$

For $t=4 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $4 s$ for t to get $d$ .

$d=(0 m/s)(4 s)+12(3.0 m/s2)(4 s)2=24 m$

For $t=5 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $5 s$ for t to get $d$ .

$d=(0 m/s)(5 s)+12(3.0 m/s2)(5 s)2=37.5 m$

The distance time graph is,

The Physics of Everyday Phenomena, Chapter 2, Problem 18E , additional homework tip 2

Figure(2)

Conclusion:

Therefore, the distances after 1s, 2s, 3s, 4s and 5s are $1.5 m$ , $6.0 m$ , $13.5 m$ , $24 m$ and $37.5 m$ respectively.

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Answer 4

Question

Chapter 2, Problem 18E

(a)

To determine

The velocity of the car after each second.

(a)

Expert Solution

Answer to Problem 18E

The velocities after 1s, 2s, 3s, 4s and 5s are $3.0 m/s$ , $6.0 m/s$ , $9.0 m/s$ , $12 m/s$ and $15 m/s$ respectively.

Explanation of Solution

Given info: The car accelerates at $3.0 m/s2$ and the time is 5 seconds.

Write the formula for average acceleration.

$a=vf−vit$

Here,

a is the average acceleration

$vf$ is the final velocity

$vi$ is the initial velocity

t is the time

Re-arrange the equation to get $vf$ .

$vf=vi+at$

For $t=1 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and 1 s for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(1 s)=3.0 m/s$

For $t=2 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $2 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(2 s)=6.0 m/s$

For $t=3 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $3 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(3 s)=9.0 m/s$

For $t=4 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $4 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(4 s)=12 m/s$

For $t=5 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $5 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(5 s)=15 m/s$

The velocity time graph is given below.

The Physics of Everyday Phenomena, Chapter 2, Problem 18E , additional homework tip 1

Figure (1)

Conclusion:

Therefore, the velocities after 1s, 2s, 3s, 4s and 5s are $3.0 m/s$ , $6.0 m/s$ , $9.0 m/s$ , $12 m/s$ and $15 m/s$ respectively.

(b)

To determine

The distance travelled after each second.

(b)

Expert Solution

Answer to Problem 18E

The distances after 1s, 2s, 3s, 4s and 5s are $1.5 m$ , $6.0 m$ , $13.5 m$ , $24 m$ and $37.5 m$ respectively.

Explanation of Solution

Given info: Acceleration is $4 m/s2$ and time is 5 seconds.

Write the formula to calculate the distance.

$d=v0t+12at2$

Here,

d is the distance

For $t=1 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $1 s$ for t to get $d$ .

$d=(0 m/s)(1 s)+12(3.0 m/s2)(1 s)2=1.5 m$

For $t=2 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $2 s$ for t to get $d$ .

$d=(0 m/s)(2 s)+12(3.0 m/s2)(2 s)2=6.0 m$

For $t=3 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $3 s$ for t to get $d$ .

$d=(0 m/s)(3 s)+12(3.0 m/s2)(3 s)2=13.5 m$

For $t=4 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $4 s$ for t to get $d$ .

$d=(0 m/s)(4 s)+12(3.0 m/s2)(4 s)2=24 m$

For $t=5 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $5 s$ for t to get $d$ .

$d=(0 m/s)(5 s)+12(3.0 m/s2)(5 s)2=37.5 m$

The distance time graph is,

The Physics of Everyday Phenomena, Chapter 2, Problem 18E , additional homework tip 2

Figure(2)

Conclusion:

Therefore, the distances after 1s, 2s, 3s, 4s and 5s are $1.5 m$ , $6.0 m$ , $13.5 m$ , $24 m$ and $37.5 m$ respectively.

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Answer 5

Chapter 2, Problem 18E

(a)

To determine

The velocity of the car after each second.

(a)

Expert Solution

Answer to Problem 18E

The velocities after 1s, 2s, 3s, 4s and 5s are $3.0 m/s$ , $6.0 m/s$ , $9.0 m/s$ , $12 m/s$ and $15 m/s$ respectively.

Explanation of Solution

Given info: The car accelerates at $3.0 m/s2$ and the time is 5 seconds.

Write the formula for average acceleration.

$a=vf−vit$

Here,

a is the average acceleration

$vf$ is the final velocity

$vi$ is the initial velocity

t is the time

Re-arrange the equation to get $vf$ .

$vf=vi+at$

For $t=1 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and 1 s for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(1 s)=3.0 m/s$

For $t=2 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $2 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(2 s)=6.0 m/s$

For $t=3 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $3 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(3 s)=9.0 m/s$

For $t=4 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $4 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(4 s)=12 m/s$

For $t=5 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $5 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(5 s)=15 m/s$

The velocity time graph is given below.

The Physics of Everyday Phenomena, Chapter 2, Problem 18E , additional homework tip 1

Figure (1)

Conclusion:

Therefore, the velocities after 1s, 2s, 3s, 4s and 5s are $3.0 m/s$ , $6.0 m/s$ , $9.0 m/s$ , $12 m/s$ and $15 m/s$ respectively.

(b)

To determine

The distance travelled after each second.

(b)

Expert Solution

Answer to Problem 18E

The distances after 1s, 2s, 3s, 4s and 5s are $1.5 m$ , $6.0 m$ , $13.5 m$ , $24 m$ and $37.5 m$ respectively.

Explanation of Solution

Given info: Acceleration is $4 m/s2$ and time is 5 seconds.

Write the formula to calculate the distance.

$d=v0t+12at2$

Here,

d is the distance

For $t=1 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $1 s$ for t to get $d$ .

$d=(0 m/s)(1 s)+12(3.0 m/s2)(1 s)2=1.5 m$

For $t=2 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $2 s$ for t to get $d$ .

$d=(0 m/s)(2 s)+12(3.0 m/s2)(2 s)2=6.0 m$

For $t=3 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $3 s$ for t to get $d$ .

$d=(0 m/s)(3 s)+12(3.0 m/s2)(3 s)2=13.5 m$

For $t=4 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $4 s$ for t to get $d$ .

$d=(0 m/s)(4 s)+12(3.0 m/s2)(4 s)2=24 m$

For $t=5 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $5 s$ for t to get $d$ .

$d=(0 m/s)(5 s)+12(3.0 m/s2)(5 s)2=37.5 m$

The distance time graph is,

The Physics of Everyday Phenomena, Chapter 2, Problem 18E , additional homework tip 2

Figure(2)

Conclusion:

Therefore, the distances after 1s, 2s, 3s, 4s and 5s are $1.5 m$ , $6.0 m$ , $13.5 m$ , $24 m$ and $37.5 m$ respectively.

Answer 6

Chapter 2, Problem 18E

(a)

To determine

The velocity of the car after each second.

(a)

Expert Solution

Answer to Problem 18E

The velocities after 1s, 2s, 3s, 4s and 5s are $3.0 m/s$ , $6.0 m/s$ , $9.0 m/s$ , $12 m/s$ and $15 m/s$ respectively.

Explanation of Solution

Given info: The car accelerates at $3.0 m/s2$ and the time is 5 seconds.

Write the formula for average acceleration.

$a=vf−vit$

Here,

a is the average acceleration

$vf$ is the final velocity

$vi$ is the initial velocity

t is the time

Re-arrange the equation to get $vf$ .

$vf=vi+at$

For $t=1 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and 1 s for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(1 s)=3.0 m/s$

For $t=2 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $2 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(2 s)=6.0 m/s$

For $t=3 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $3 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(3 s)=9.0 m/s$

For $t=4 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $4 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(4 s)=12 m/s$

For $t=5 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $5 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(5 s)=15 m/s$

The velocity time graph is given below.

The Physics of Everyday Phenomena, Chapter 2, Problem 18E , additional homework tip 1

Figure (1)

Conclusion:

Therefore, the velocities after 1s, 2s, 3s, 4s and 5s are $3.0 m/s$ , $6.0 m/s$ , $9.0 m/s$ , $12 m/s$ and $15 m/s$ respectively.

Answer 7

Chapter 2, Problem 18E

(a)

To determine

The velocity of the car after each second.

Answer 8

(a)

Expert Solution

Answer to Problem 18E

The velocities after 1s, 2s, 3s, 4s and 5s are $3.0 m/s$ , $6.0 m/s$ , $9.0 m/s$ , $12 m/s$ and $15 m/s$ respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info: The car accelerates at $3.0 m/s2$ and the time is 5 seconds.

Write the formula for average acceleration.

$a=vf−vit$

Here,

a is the average acceleration

$vf$ is the final velocity

$vi$ is the initial velocity

t is the time

Re-arrange the equation to get $vf$ .

$vf=vi+at$

For $t=1 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and 1 s for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(1 s)=3.0 m/s$

For $t=2 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $2 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(2 s)=6.0 m/s$

For $t=3 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $3 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(3 s)=9.0 m/s$

For $t=4 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $4 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(4 s)=12 m/s$

For $t=5 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $5 s$ for t to get $vf$ .

$vf=(0 m/s)+(3.0 m/s2)(5 s)=15 m/s$

The velocity time graph is given below.

The Physics of Everyday Phenomena, Chapter 2, Problem 18E , additional homework tip 1

Figure (1)

Conclusion:

Therefore, the velocities after 1s, 2s, 3s, 4s and 5s are $3.0 m/s$ , $6.0 m/s$ , $9.0 m/s$ , $12 m/s$ and $15 m/s$ respectively.

Answer 13

(b)

To determine

The distance travelled after each second.

(b)

Expert Solution

Answer to Problem 18E

The distances after 1s, 2s, 3s, 4s and 5s are $1.5 m$ , $6.0 m$ , $13.5 m$ , $24 m$ and $37.5 m$ respectively.

Explanation of Solution

Given info: Acceleration is $4 m/s2$ and time is 5 seconds.

Write the formula to calculate the distance.

$d=v0t+12at2$

Here,

d is the distance

For $t=1 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $1 s$ for t to get $d$ .

$d=(0 m/s)(1 s)+12(3.0 m/s2)(1 s)2=1.5 m$

For $t=2 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $2 s$ for t to get $d$ .

$d=(0 m/s)(2 s)+12(3.0 m/s2)(2 s)2=6.0 m$

For $t=3 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $3 s$ for t to get $d$ .

$d=(0 m/s)(3 s)+12(3.0 m/s2)(3 s)2=13.5 m$

For $t=4 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $4 s$ for t to get $d$ .

$d=(0 m/s)(4 s)+12(3.0 m/s2)(4 s)2=24 m$

For $t=5 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $5 s$ for t to get $d$ .

$d=(0 m/s)(5 s)+12(3.0 m/s2)(5 s)2=37.5 m$

The distance time graph is,

The Physics of Everyday Phenomena, Chapter 2, Problem 18E , additional homework tip 2

Figure(2)

Conclusion:

Therefore, the distances after 1s, 2s, 3s, 4s and 5s are $1.5 m$ , $6.0 m$ , $13.5 m$ , $24 m$ and $37.5 m$ respectively.

Answer 14

(b)

To determine

The distance travelled after each second.

Answer 15

(b)

Expert Solution

Answer to Problem 18E

The distances after 1s, 2s, 3s, 4s and 5s are $1.5 m$ , $6.0 m$ , $13.5 m$ , $24 m$ and $37.5 m$ respectively.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given info: Acceleration is $4 m/s2$ and time is 5 seconds.

Write the formula to calculate the distance.

$d=v0t+12at2$

Here,

d is the distance

For $t=1 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $1 s$ for t to get $d$ .

$d=(0 m/s)(1 s)+12(3.0 m/s2)(1 s)2=1.5 m$

For $t=2 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $2 s$ for t to get $d$ .

$d=(0 m/s)(2 s)+12(3.0 m/s2)(2 s)2=6.0 m$

For $t=3 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $3 s$ for t to get $d$ .

$d=(0 m/s)(3 s)+12(3.0 m/s2)(3 s)2=13.5 m$

For $t=4 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $4 s$ for t to get $d$ .

$d=(0 m/s)(4 s)+12(3.0 m/s2)(4 s)2=24 m$

For $t=5 s$ ,

Substitute 0 m/s for $vi$ , $3.0 m/s2$ for a and $5 s$ for t to get $d$ .

$d=(0 m/s)(5 s)+12(3.0 m/s2)(5 s)2=37.5 m$

The distance time graph is,

The Physics of Everyday Phenomena, Chapter 2, Problem 18E , additional homework tip 2

Figure(2)

Conclusion:

Therefore, the distances after 1s, 2s, 3s, 4s and 5s are $1.5 m$ , $6.0 m$ , $13.5 m$ , $24 m$ and $37.5 m$ respectively.

Answer 20

Ch. 2 - Prob. 1CQ Ch. 2 - Suppose we choose inches as our basic unit of...Ch. 2 - What units would have an appropriate size for...Ch. 2 - A tortoise and a hare cover the same distance in a...Ch. 2 - A driver states that she was doing 80 when stopped...Ch. 2 - Does the speedometer on a car measure average...Ch. 2 - Is the average speed over several minutes more...Ch. 2 - The highway patrol sometimes uses radar guns to...Ch. 2 - Is the term vehicle density (as used in everyday...Ch. 2 - Prob. 10CQ

The velocity of the car after each second.

The velocity of the car after each second.

Answer to Problem 18E

Explanation of Solution

The distance travelled after each second.

Answer to Problem 18E

Explanation of Solution

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