Question

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

Answer 1

Answer:

240 meters

Explanation:

The distance traveled by the vehicle can be calculated using the following equation:

$v_{f}^{2} = v_{0}^{2} + 2ax$   (1)

Where:

x: is the displacement

$v_{f}$: is the final speed = 0 (reduces its velocity back to zero)                    

$v_{0}$: is the initial speed = 60 m/s

a: is the acceleration = -7.5 m/s²

By solving equation (1) for x we have:

$x = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{0 - (60 m/s)^{2}}{2*(-7.5 m/s^{2})} = 240 m$

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

           

I hope it helps you!