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Calculating Final Velocity and Stopping Time

Created by @nathanedwards
 at October 31st 2023, 5:48:38 am.
AP_Physics_1-Questions
#motion
#physics
#velocity

Question:

A car starts from rest and accelerates at a constant rate of 4 m/s^2 for a distance of 100 meters.

a) Calculate the final velocity of the car.

b) How long does it take for the car to reach its final velocity?

c) If the car then decelerates at a constant rate of 2 m/s^2 for 50 meters, calculate its final velocity.

d) How long does it take for the car to come to a stop?

Answer:

a) To calculate the final velocity of the car, we can use the equation:

v2=u2+2asv^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity (resting in this case), a is the acceleration, and s is the distance traveled.

Given that the initial velocity is 0 m/s, the acceleration is 4 m/s^2, and the distance traveled is 100 meters, we can substitute these values into the equation:

v2=02+2⋅4⋅100v^2 = 0^2 + 2 \cdot 4 \cdot 100
v2=0+800v^2 = 0 + 800
v2=800v^2 = 800

Taking the square root of both sides, we find:

v=800≈28.28 m/sv = \sqrt{800} \approx 28.28\,\text{m/s}

Therefore, the final velocity of the car is approximately 28.28 m/s.

b) To calculate the time it takes for the car to reach its final velocity, we can use the equation:

v=u+atv = u + at

where t is the time taken.

Given that the initial velocity is 0 m/s and the final velocity is 28.28 m/s, we can substitute these values into the equation:

28.28=0+4t28.28 = 0 + 4t

Solving for t, we get:

t=28.284=7.07 st = \frac{28.28}{4} = 7.07\,\text{s}

Therefore, it takes approximately 7.07 seconds for the car to reach its final velocity.

c) To calculate the final velocity of the car after deceleration, we can use the same equation as in part a:

v2=u2+2asv^2 = u^2 + 2as

Given that the initial velocity is 28.28 m/s (from part a), the deceleration is 2 m/s^2, and the distance traveled is 50 meters, we can substitute these values into the equation:

v2=(28.28)2+2⋅(−2)⋅50v^2 = (28.28)^2 + 2 \cdot (-2) \cdot 50
v2=800−200v^2 = 800 - 200
v2=600v^2 = 600

Taking the square root of both sides, we find:

v=600≈24.5 m/sv = \sqrt{600} \approx 24.5\,\text{m/s}

Therefore, the final velocity of the car after deceleration is approximately 24.5 m/s.

d) To calculate the time it takes for the car to come to a stop, we can use the equation:

v=u+atv = u + at

where t is the time taken.

Given that the initial velocity is 24.5 m/s (from part c), the final velocity is 0 m/s, and the deceleration is -2 m/s^2, we can substitute these values into the equation:

0=24.5+(−2)t0 = 24.5 + (-2)t

Solving for t, we get:

−2t=−24.5-2t = -24.5
t=−24.5−2=12.25 st = \frac{-24.5}{-2} = 12.25\,\text{s}

Therefore, it takes approximately 12.25 seconds for the car to come to a stop.

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