When a drag strip vehicle reaches a velocity of 60 m/s, it begins a negative acceleration by releasing a drag chute and applying its brakes. While reducing its velocity back to zero, its acceleration along a straight line path is a constant -7.5 m/s2. What displacement does it undergo during this deceleration period?

I think I am plugging the problems in wrong I am not getting the correct answer?

Vf^2=Vi^2+2ad solve for d.

How can I know how you are "plugging"?

"Plug" this into the google search window:

60^2/(2*7.5)=

That worked. Thank you

To calculate the displacement during the deceleration period, we need to use the equations of motion. In this case, we know the initial velocity (u), final velocity (v), and acceleration (a) of the vehicle.

Given:
Initial velocity, u = 60 m/s
Final velocity, v = 0 m/s
Acceleration, a = -7.5 m/s^2 (negative because it is decelerating)

We can use the kinematic equation: v^2 = u^2 + 2as, where s represents displacement.

Rearranging the equation to solve for displacement:
s = (v^2 - u^2) / (2a)

Substituting the known values:
s = (0^2 - 60^2) / (2 * -7.5)

Now we can calculate the displacement:

s = (-3600) / (-15)
s = 240 meters

Therefore, the vehicle undergoes a displacement of 240 meters during the deceleration period.