Engineering Mechanics: Dynamics Study (Book and Pearson eText)
Engineering Mechanics: Dynamics Study (Book and Pearson eText)
14th Edition
ISBN: 9780134116990
Author: Russell C. Hibbeler
Publisher: PEARSON
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Chapter 12.2, Problem 1PP

a. If s = (2t3) m, where t is in seconds, determine v when t = 2 s.

b. If v = (5s) m/s, where s is in meters, determine a at s = 1 m.

c. If v = (4t + 5) m/s, where t is in seconds, determine a when t = 2 s.

d. If a = 2 m/s2, determine v when t = 2 s if v = 0 when t = 0.

e. If a = 2 m/s2, determine v at s= 4 m if v = 3 m/s at s = 0.

f. If a = (s) m/s2, where s is in meters, determine v when s = 5 m if v = 0 at s = 4 m

g. If a = 4 m/s2, determine s when t = 3 s if v = 2 m/s and s = 2 m when t = 0.

h. It a = (8t2) m/s2, determine v when t = 1 s if v = 0 at t = 0.

i. If s = (3t2 + 2) m, determine v when t = 2 s.

j. When t = 0 the particles is at A. In four seconds it travels to B, then in another six seconds it travels to C. Determine the average velocity and the average speed. The origin of the coordinate is at O.

Chapter 12.2, Problem 1PP, a. If s = (2t3) m, where t is in seconds, determine v when t = 2 s. b. If v = (5s) m/s, where s is

a)

Expert Solution
Check Mark
To determine

The velocity when time is t=2s .

Answer to Problem 1PP

The velocity when time is t=2s is 24m/s .

Explanation of Solution

Given:

The time is t=2s .

The distance equation is s=(2t3)m .

Write the distance equation.

s=(2t3)m

Here, average velocity is υavg , change in distance is Δs and change in time is Δt .

Write the expression velocity.

υ=dsdt (I).

Here, velocity is υ and rate of change of distance s with respect to time (t) is dsdt .

Conclusion:

Substitute (2t3)m for s in Equation (I).

υ=ddt(2t3)=6t2m/s (II).

Substitute 2s for t in Equation (II).

υ=6t2m/s=6(2)2=24m/s

Thus, the velocity when time is t=2s is 24m/s .

b)

Expert Solution
Check Mark
To determine

The acceleration when distance s=1m .

Explanation of Solution

The acceleration a when distance s=1m 25m/s2 .

Given:

The distance is s=1m .

The velocity equation is υ=(5s)m/s .

Write the velocity equation.

υ=(5s)m/s

Write the expression acceleration.

a=υdυds (I).

Here, velocity is υ , acceleration is a and rate of change of velocity υ with respect to distance (s) is dυds .

Conclusion:

Substitute (5s)m/s for υ in Equation (I).

a=υdυds=(5s)dds(5s)=5s(5)=25s (II).

Substitute 1m for s in Equation (II).

a=25s=25(1)=25m/s2

Thus, the acceleration a when distance s=1m is 25m/s2 .

c)

Expert Solution
Check Mark
To determine

The acceleration when distance t=2s .

Answer to Problem 1PP

The acceleration a when distance t=2s is 4m/s2 .

Explanation of Solution

Given:

The distance is s=1m .

The velocity equation is υ=(4t+5)m/s .

Write the velocity equation.

υ=(4t+5)m/s

Write the expression acceleration.

a=dυdt (I).

Here, acceleration is a and rate of change of velocity υ with respect to time (t) is dυdt .

Conclusion:

Substitute (4t+5)m/s for υ in Equation (I).

a=dυdt=ddt(4t+5)=4m/s (II).

Thus, the acceleration a when distance t=2s is 4m/s2 .

d)

Expert Solution
Check Mark
To determine

The velocity when time is t=2s .

Answer to Problem 1PP

The velocity υ when time is t=2s is 4m/s .

Explanation of Solution

Given:

The time is t=2s .

The acceleration is a=2m/s2 .

The initial velocity is υ0=0

Write the expression for final velocity in y direction.

υ=υ0+act (III).

Here, final velocity is υ , initial velocity is υ0 , acceleration due to gravity is ac and time is t .

Conclusion:

Substitute 2s for t , 0 for υ0 (s0) and 2m/s2 for ac in Equation (I).

υ=υ0+act=0+(2)(2)=4m/s

Thus, the velocity υ when time is t=2s is 4m/s .

e)

Expert Solution
Check Mark
To determine

The velocity when distance is s=4m .

Answer to Problem 1PP

The velocity υ when distance is s=4m is 5m/s .

Explanation of Solution

Given:

The time is t=3s .

The acceleration is a=2m/s2 .

The initial velocity is υ0=3m/s

The initial distance is (s0)=0 .

The final distance is s=4m .

Write the expression for final velocity in y direction..

υ2=(υ02)+2ac(sy(s0)) (II).

Here, final velocity is υ , initial velocity is (υ0)y , acceleration due to gravity is ac , final distance is s , initial distance is (s0)y .

Conclusion:

Substitute 3m/s for υ0 , 4m for s , 0 for (s0) and 2m/s2 for ay in Equation (I).

υ2=(υ02)+2ac(s(s0))υ2=(3)2+2(2)(40)υ=5m/s

Thus, the velocity υ when distance is s=4m is 5m/s .

f)

Expert Solution
Check Mark
To determine

The velocity

Answer to Problem 1PP

The velocity υ is 3m/s .

Explanation of Solution

Given:

The distance is s1=5m .

The distance is s2=4m .

The acceleration equation is a=(s)m/s2 .

Write the acceleration equation.

a=(s)m/s2

Write the expression acceleration.

a=υdυdsads=υdυ (I).

Here, velocity is υ , acceleration is a and rate of change of velocity υ with respect to distance (s) is dυds .

Conclusion:

Substitute (s)m/s2 for a in Equation (I).

Integrate the Equation (I) at the limits 0 to υ for υ and s1=5m to s2=4m for s .

0υdυ=45(s)ds[υ22]0υ=[s22]45

υ22=(5)22(4)22υ2=2516υ=3m/s

Thus, the velocity υ is 3m/s .

g)

Expert Solution
Check Mark
To determine

The distance when time is t=3s .

Answer to Problem 1PP

The distance s when time is t=3s is 26m .

Explanation of Solution

Given:

The time is t=3s .

The acceleration is a=4m/s2 .

The velocity is υ=2m/s

The distance is (s0)=2m

Write the expression for final distance in y direction.

s=(s0)+(υ)t+12at2 (I).

Here, final distance is s , initial distance is (s0) , initial velocity is (υ) , acceleration due to gravity is a and time is t .

Conclusion:

Substitute 3s for t , 2m/s for υ , 2m for (s0) and 4m/s2 for ay in Equation (I).

s=(s0)+(υ)t+12at2=2+2(3)+12(4)(3)2=26m

Thus, the distance s when time is t=3s is 26m .

h)

Expert Solution
Check Mark
To determine

The velocity when time is t=1s .

Answer to Problem 1PP

The velocity υ when time is t=1s is 2.67m/s .

Explanation of Solution

Given:

The time is t=1s .

The acceleration equation is a=(8t2)m/s2 .

Write the acceleration equation.

a=(8t2)m/s2

Write the expression acceleration.

a=dυdtadt=dυ (I).

Here, acceleration is a and rate of change of velocity υ with respect to time (t) is dυdt .

Conclusion:

Substitute (8t2)m/s2 for a in Equation (I).

Integrate the Equation (I) at the limits 0 to υ for υ and 0 to t for t .

0υdυ=0t(8t2)dt[υ]0υ=[8t33]0tυ=(2.667t3)m/s (II).

Substitute 2s for t in Equation (II).

υ=(2.667t3)m/s=2.667(1)=2.67m/s

Thus, the velocity when time is t=1s is 2.67m/s .

i)

Expert Solution
Check Mark
To determine

The velocity when time is t=2s .

Answer to Problem 1PP

The velocity υ when time is t=2s is 12m/s .

Explanation of Solution

Given:

The time is t=2s .

The distance equation is s=(3t2+2)m .

Write the distance equation.

s=(3t2+2)m

Here, average velocity is υavg , change in distance is Δs and change in time is Δt .

Write the expression velocity.

υ=dsdt (I).

Here, velocity is υ and rate of change of distance s with respect to time (t) is dsdt .

Conclusion:

Substitute (3t2+2)m for s in Equation (I).

υ=ddt(3t2+2)=6tm/s (II).

Substitute 2s for t in Equation (II).

υ=6tm/s=6(2)=12m/s

Thus, the velocity υ when time is t=2s is 12m/s .

j)

Expert Solution
Check Mark
To determine

The average velocity and the average speed of the particle.

Answer to Problem 1PP

The average velocity of particle is 0.7m/s .

The average speed of particle is 2.1m/s .

Explanation of Solution

Given:

The distance traveled by the particle from A to C is shown in Figure (1).

Engineering Mechanics: Dynamics Study (Book and Pearson eText), Chapter 12.2, Problem 1PP

The time traveled by the particle from A to B is 4s .

The time traveled by the particle from B to C is 6s .

Write the expression for the average velocity.

υavg=ΔsΔt (I).

Here, average velocity is υavg , change in distance is Δs and change in time is Δt .

Write the expression for the average speed υsp .

υsp=sTotaltTotal (II).

Here, the total distance is sTotal and the total time traveled by the particle is tTotal .

Refer Figure (1) and calculate the total distance traveled by the particle.

sTotal=sAB+sBC (III)

Refer Figure (1) and calculate the total time traveled by the particle.

tTotal=tAB+tBC (IV)

Conclusion:

From the Figure (1) calculate the change in distance.

Δs=(s2s1)=(6(1))=7m

Calculate the change in distance

Δt=(t2t1)=(100)=10s

Substitute 10s for Δt and 7m for Δs in Equation (1).

υavg=ΔsΔt=7m10s=0.7m/s

Thus, the average velocity of particle is 0.7m/s .

The time traveled by the particle from A to B .

tAB=6s

The time traveled by the particle from B to C .

tBC=4s

Substitute 7m for sAB and 14m for sBC in Equation (III).

sTotal=sAB+sBC=7m+14m=21m

Substitute 6s for tAB and 4s for tBC in Equation (IV).

ttotal=6s+4s=10s

Substitute 21m for sTotal and 10s for tTotal in Equation (II).

υsp=sTotaltTotal=21m10s=2.1m/s

Thus, the average speed of particle is 2.1m/s

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Chapter 12 Solutions

Engineering Mechanics: Dynamics Study (Book and Pearson eText)

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such that its position as a function of time is...Ch. 12.8 - Determine the radial and transverse components of...Ch. 12.8 - Determine the magnitudes of the velocity and...Ch. 12.8 - Determine the velocity and acceleration of the...Ch. 12.8 - Determine the radial and transverse components of...Ch. 12.8 - If it is assumed that the hose lies in a...Ch. 12.8 - Two pin-connected slider blocks, located at B....Ch. 12.8 - Determine the magnitude of the acceleration of the...Ch. 12.8 - If the geometry of the fixed rod for a short...Ch. 12.8 - The platform rotates at a constant rate of 6...Ch. 12.8 - Determine the cars radial and transverse...Ch. 12.8 - Determine the cars radial and transverse...Ch. 12.8 - If it maintains a constant speed of v = 35 ft/s,...Ch. 12.8 - Determine the cylindrical components of the...Ch. 12.8 - Determine the maximum and minimum magnitudes of...Ch. 12.8 - The peg is constrained to move in the slots of the...Ch. 12.8 - When = 30, the angular velocity and angular...Ch. 12.8 - Determine the angular rate of rotation of the...Ch. 12.8 - A truck is traveling along the horizontal circular...Ch. 12.8 - Two pin-connected slider blocks, located at B,...Ch. 12.8 - Determine the magnitude of the acceleration of the...Ch. 12.8 - The searchlight on the boat anchored 2000 ft from...Ch. 12.8 - If the car in Prob.12-187 is accelerating at 15...Ch. 12.8 - If = 4 rad/s (constant), determine the radial and...Ch. 12.8 - if the particle has an angular acceleration = 5...Ch. 12.8 - If = (0.5t)rad, where t is in seconds, determine...Ch. 12.8 - Determine the magnitudes of the velocity and...Ch. 12.8 - When t = 0, = 0. Use Simpson's rule with n = 50...Ch. 12.8 - The double collar C is pin connected together such...Ch. 12.10 - Determine the velocity of block D if end A of the...Ch. 12.10 - Determine the velocity of block A if end B of the...Ch. 12.10 - Determine the velocity of block A if end B of the...Ch. 12.10 - Determine the velocity of block A if end F of the...Ch. 12.10 - Determine the velocity of car A if point P on the...Ch. 12.10 - Determine the velocity of cylinder B if cylinder A...Ch. 12.10 - Determine the velocity of car B relative to car A.Ch. 12.10 - Determine the magnitude and direction of the...Ch. 12.10 - Determine the distance between them when t = 4 s.Ch. 12.10 - If B is accelerating at 1200 km/h2 while A...Ch. 12.10 - If the end of the cable at A is pulled down with a...Ch. 12.10 - The motor at D draws in its cable at aD = 5 m/s2....Ch. 12.10 - If BC remains fixed while the plunger P is pushed...Ch. 12.10 - If the end of the cable at A is pulled down with a...Ch. 12.10 - Determine the displacement of the log if the truck...Ch. 12.10 - Determine the constant speed at which the cable at...Ch. 12.10 - Determine the time needed to lift the load 7 m.Ch. 12.10 - If the end A of the cable is moving at vA = 3 m/s,...Ch. 12.10 - Determine the time needed for the load at B to...Ch. 12.10 - Determine the velocity of the block.Ch. 12.10 - If block A of the pulley system is moving downward...Ch. 12.10 - Determine the speed of the block at B.Ch. 12.10 - Determine the speed of block A if the end of the...Ch. 12.10 - The motor draws in the cable at D with a constant...Ch. 12.10 - The pulley at A is attached to the smooth collar...Ch. 12.10 - When sB = 6ft. the end of the cord at B is pulled...Ch. 12.10 - Determine the velocity and acceleration of block B...Ch. 12.10 - Determine how fast the boat approaches the pier at...Ch. 12.10 - If the hydraulic cylinder H draws in rod BC at 2...Ch. 12.10 - The car at B is traveling at 18.5 m/s along the...Ch. 12.10 - When sA = 1.5 m, vB = 6 m/s. Determine the...Ch. 12.10 - If block B is moving down with a velocity vB and...Ch. 12.10 - Determine the velocity and acceleration of the...Ch. 12.10 - If their velocities are vA = 500km/h and vB =...Ch. 12.10 - If B is increasing its speed by 1200mi/h2, while A...Ch. 12.10 - The point of destination is located along the...Ch. 12.10 - If vA = 40ft/s and vB = 30 ft/s. determine the...Ch. 12.10 - An instrument in the car indicates that the wind...Ch. 12.10 - If vA = 10m/s and vB = 15m/s, determine the...Ch. 12.10 - At the same instant, car B is decelerating at 250...Ch. 12.10 - At the instant shown, A has a speed of 90ft/sand...Ch. 12.10 - If raindrops fall vertically at 7 km/h in still...Ch. 12.10 - If B is increasing its velocity by 2 m/s2, while A...Ch. 12.10 - If A is increasing its speed at 4 m/s2, whereas...Ch. 12.10 - Compute the terminal (constant) velocity vr of the...Ch. 12.10 - He wishes to cross the 40-ft-wide river to point...Ch. 12.10 - Determine the magnitude and direction of the...Ch. 12.10 - At the instant the ball is thrown, the player is...Ch. 12.10 - At the instant the ball is thrown, the player is...Ch. 12.10 - Determine the constant speed at which the player...Ch. 12.10 - At this same instant car B travels along the...Ch. 12.10 - If you measured the time it takes for the...Ch. 12.10 - Determine its maximum acceleration and maximum...Ch. 12.10 - Originally s0 = 0.Ch. 12.10 - A projectile, initially at the origin, moves along...Ch. 12.10 - Determine the acceleration when t = 2.5 s, 10 s,...Ch. 12.10 - If it takes 3 s to go from A to B, and then 5 s to...Ch. 12.10 - From a videotape, it was observed that a player...Ch. 12.10 - The truck travels in a circular path having a...Ch. 12.10 - If the car starts from rest when = 0, determine...Ch. 12.10 - Determine the magnitude of the particles...Ch. 12.10 - Determine the time needed for the load at B to...Ch. 12.10 - If their velocities are vA = 600 km/h and vB = 500...
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