A child appears to be running into the street aheadIt takes 23 seconds for the driver to react and begin to brake, but this time at a rate of - 7.5 m/s ^ 2 . What is the stopping distance for the car in this situation?

Explanation:

Given

acceleration a = - 7.5 m/s^2

time t = 23secs

Required

Stopping distance S

First let us get the initial velocity if the child using:

v = u +  at

Note that v = 0

0 = u -7.5(23)

u = 7.5 * 23

u =172.5m/s

Using the equation of motion to get the stopping distance:

v² = u²+2aS

0 = u²+2aS

-u² = 2aS

-172.5² = 2(-7.5)S

29,756.25 = 15S

S =29,756.25/15

S = 1,983.75m

Hence the stopping distance is 1,983.75m

the independent variable, or the one that is intentionally manipulated is the speed.

6 cos 34, 6 sin 34 trig