Car Velocity Question

In summary, part 1 introduces a scenario where a car is traveling with constant acceleration and has to cover a distance of 81.4 m between two checkpoints within 5.89 s. Part 2 asks to find the velocity at checkpoint B when the acceleration is 7.45 m/s^2. Some attempts are made to solve this problem, including using the formula d = Vi*t + 0.5a*t^2.
  • #1

Homework Statement

Part 1 Consider a car which is traveling along a straight road with constant acceleration a. There are two checkpoints, A and B , which are a distance 81.4 m apart. The time it takes for the car to travel from A to B is 5.89 s.

Part 2 Find the velocity VB for the case where the acceleration is 7.45 m/s^2.

The Attempt at a Solution

acceleration multiplied by the final time does not give me the answer
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  • #2
What is VB? Is that supposed to be the velocity once it reaches checkpoint B?
  • #3
yea my browser won't load the functions so it should read V(sub)B which is the velocity once it reaches checkpoint B.
  • #4
Perhaps start by finding what the velocity at checkpoint A is.
  • #5
how would you go by doing that?
  • #6
anyone? I have no clue how to find the velocity at either checkpoints with the given data
  • #7
Take A to be the initial point, time = 0.
Then for the trip to B you know d, t, and a. Don't know Vi.
You could make progress with d = Vi*t + 0.5a*t^2 which has only one unknown.
There is another formula like that with a Vf instead of a Vi which would get you to the answer in one step.

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