# Drag racer velocity and acceleration

## Homework Statement

A drag racer, starting from rest, speeds up for 402 m with an acceleration of +23.0 m/s2. A parachute then opens, slowing the car down with an acceleration of -6.70 m/s2. How fast is the racer moving 3.35 102 m after the parachute opens?

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## The Attempt at a Solution

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G01
Homework Helper
Gold Member
In order to get help here you have to show some work. It's the rules. You can't possibly know NOTHING about this problem. Give me your thoughts at least. What equations will be involved? What principles are involved?

It is clearly a kinematics problem, so I am guessing it has to do with acceleration and velocity. All the variables are given in terms of acceleration and the final answer has to be in terms of velocity, so I assume the starting equation will be: vf = vi + a*t. The only thing is that any of the equations I look at, they are all missing two or more variables.

G01
Homework Helper
Gold Member
Ok, now do you know the time the parachute opens or is this not given?

If your not given the time, I suggest looking for a different equation.

It is not given. It does not give initial velocity either.

G01
Homework Helper
Gold Member
Ok, then, can you find the time that the car reaches the 402m mark from the information given?

EDIT: Also, if the car starts at rest, what is the initial velocity?

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