The driver of a car wishes to pass a truck that is traveling at a constant speed

Question

The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.1 m/s . Initially, the car is also traveling at a speed 20.1 m/s and its front bumper is a distance 23.3 mbehind the truck's rear bumper. The car begins accelerating at a constant acceleration 0.550 m/s2 , then pulls back into the truck's lane when the rear of the car is a distance 25.9 m ahead of the front of the truck. The car is of length 5.00 m and the truck is of length 22.0 m .

Part A

How much time is required for the car to pass the truck?

Part B

What distance does the car travel during this time?

Part C

What is the final speed of the car?

Explanation / Answer

part A

let's make it simple and put the frame of motion traveling with the truck at 20.1 m/s. Motion is relative, and part A is asking us just for the time, so it's fine.

The car needs to move 23.3 m (current distance from truck) + 5.00 m (length of car since we started measuring from front bumper and ended at rear bumper) + 25.9 m (same thing with the truck but rear->front) + 26.6m (distance between vehicles at end) = 80.8m

So the car needs to move 80.8m after accelerating at 0.550 m/s^2 from rest. This can be solved with

displacement = (1/2) * acceleration * time^2 + 0 (since we started from rest)
80.8 = (1/2) * 0.550 * t^2
t ~17.14 seconds

Part B

Now that we know how long the process takes, we go back to using the 20.1 m/s as initial velocity to find the distance traveled. Logically speaking, the first part of the equation will be the same because we just solved it above as the acceleration portion, so we can just add 80.8m and (20.1m/s * t) to get 506m.

Part C

final velocity = acceleration * time + initial velocity
v = at + v(0)
v = (0.550 m/s^2)*(17.14s) + 20.1m/s
v = 28.63 m/s ~ 29.52 m/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.