Question

An Indy car racer traveling at 15.0 meters per second slows as he makes a pit stop. If he decelerates at 2.5 meters per second squared, he was

away from the pit when he started braking.
meters

Answer 1

x = 45 [m]

Step-by-step explanation:

To solve this problem we must use the following kinematics equation:

`v_(f)^(2) = v_(i)^(2)-(2*a*x)`

where:

Vf = final velocity = 0

Vi = initial velocity = 15 [m/s]

a = desacceleration = 2.5 [m/s^2]

x = distance [m]

Note: the negative sign in the equation above, is because the car slows down.

0^2 = 15^2 - (2*2.5*x)

2*2.5*x = 15^2

x = 45 [m]