# Car Velocity Question

• bns1201
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.

## 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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What is VB? Is that supposed to be the velocity once it reaches checkpoint B?

yea my browser won't load the functions so it should read V(sub)B which is the velocity once it reaches checkpoint B.

Perhaps start by finding what the velocity at checkpoint A is.

how would you go by doing that?

anyone? I have no clue how to find the velocity at either checkpoints with the given data

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.