Finding displacement of a drag car stopping

[CL1349]

Homework Statement

A drag racer, starting from rest, accelerates to ¼ mile with a constant acceleration whose magnitude is 12.427 m/s2. At the end of the ¼ mile, a parachute opens, slowing the car down and bringing the car to a stop after 40 seconds. From the time the parachute opened, how far did the race car travel?

Homework Equations

Vf = Vi + AT

The Attempt at a Solution

I split the problem into two parts. The first 1/4, and the 3/4 section. With only the acceleration and initial velocity given, I'm kinda lost as to what I should find. I tried to find the final velocity before the parachute opens by doing this:

(0 m/s) + (12.427 m/s2)( t ) = Vf

I got 12.427 m/s^2 (t) as the final velocity.

Then I tried to plug that velocity into the second part as the initial velocity.

A = (Vf - Vi) / T

A = (0 m/s) - (12.427 m/s^2) T / T

I canceled out the time.

Now I'm trying to figure out what to do with

0 m/s - 12.427 m/s^2 = a

I don't think I did it right. It's really confusing and frustrating me. Can anyone lead me in the right direction?

 

[Delphi51]

Welcome to PF.
The two times may be different; you can't cancel them.
For the first quarter mile, you know the distance and the acceleration.
From that, you should be able to find out anything you would like to know about that part of the motion - time, final velocity, etc. But you must use both the known acceleration and distance. Look through your list of formulas for one that has just the two knowns and something you would like to find, perhaps time or final velocity.

The final velocity for the first part is the initial velocity for the deceleration. Make sure you find it before going on to the deceleration part.

 

[CL1349]

Thanks for the reply. I see what I did wrong now.

So I converted the 1/4 mile to 402.34 meters.

I found the final velocity of the first segment using $$v^2 = v_0^2 + 2 a \Delta x$$.

The final velocity was 100 m/s for the first segment.

100 m/s is the initial velocity for the other segment. I used $$x = x_0 + v_0 t + (1/2) a t^2$$

Vi = 100 m/s
T = 40 s
Vv = 0 m/s
a = -2/5 m/s^2

What I got for the displacement was 2000 m. Is that right?

 

[Delphi51]

Looks good! That must be a typo on the acceleration, -2.5 rather than -2/5.

It is so important to know what carries forward from one part to another.

 

[rwisz]

I would approach this problem by first identifying the key variables and equations that relate to the given information. In this case, we have the acceleration (a = 12.427 m/s^2) and the time (t = 40 seconds) for the second part of the race, when the parachute is opened and the car comes to a stop. We also know that the initial velocity (Vi) is 12.427 m/s as the car starts from rest.

To find the displacement (d) of the car during this second part of the race, we can use the equation d = Vi*t + 1/2*a*t^2. Plugging in the values, we get d = (12.427 m/s)*(40 s) + 1/2*(12.427 m/s^2)*(40 s)^2 = 2490.8 meters.

Therefore, the race car traveled 2490.8 meters during the second part of the race, when the parachute was opened and it came to a stop.