A drag racer starts from rest and accelerates uniformly for 5s covering a distance of 120 m.

a. what is the car's acceleration?
I did DI=vit+1/2at^2
120=0(5)+(.5)(5^2)a
120/12.5=12.5a/12.5
a= 9.6 m/s^2

b. what is the car's final velocity at the end of the 5s?
I did vf=vi+at
vf=0+9.6(5)
vf=48 m/s

c. at the end of the 120m track the driver opens a parachute and slams on his brakes. If he only has 30 m to bring his car to rest what acceleration will he experience? *****this part I'm lost at*****

appreciate the help!

To find the acceleration experienced by the car when it is brought to rest in a distance of 30 m, you can use the following equation:

vf^2 = vi^2 + 2ad

where vf is the final velocity (0 m/s as the car comes to rest), vi is the initial velocity (which we need to find), a is the acceleration, and d is the distance traveled.

Rearranging the equation to solve for acceleration:

a = (vf^2 - vi^2) / (2d)

Now, we can substitute the known values:

vi = ? (initial velocity before braking, but we need to find it)
vf = 0 m/s (final velocity, as the car comes to rest)
d = 30 m (distance traveled in braking)

The equation becomes:

a = (0^2 - vi^2) / (2 * 30)

Simplifying further:

a = - vi^2 / 60

We can see that the acceleration is negative because it acts in the opposite direction of motion (opposite to the initial velocity).

To find the value of vi, we can use the information given in the problem that the car covers a distance of 120 m in 5 seconds. Since we already found the car's acceleration in part (a) to be 9.6 m/s^2, we can now determine the initial velocity using the equation:

d = vit + 1/2at^2

Substituting the known values:

120 = vi(5) + (1/2)(9.6)(5^2)

Simplifying:

120 = 5vi + 12(5)
120 = 5vi + 60
5vi = 120 - 60
5vi = 60
vi = 60 / 5
vi = 12 m/s

Now that we have the initial velocity (vi = 12 m/s), we can substitute it into the equation for acceleration:

a = - vi^2 / 60
a = - (12^2) / 60
a = - 144 / 60
a = - 2.4 m/s^2

Therefore, the car will experience an acceleration of -2.4 m/s^2 during braking.