Question

Determine the stopping distance for a car traveling at 16 m/s for an acceleration of -4m/s^2

Answer 1

The distance d traveled by an object under uniformly accelerated motion while it changes its speed from v_o to v_f with an acceleration a is given by the formula:

`d=(v^2_f-v^2_0)/(2a)`

Since the car travels initially at 16m/s and it stops, then v_0=16m/s and v_f=0. Replace those values into the formula, as well as a=-4m/s^2 to find the stopping distance for the car:

`d=((0)^2-(16(m)/(s))^2)/(2(-4(m)/(s^2)))=(-256(m^2)/(s^2))/(-8(m)/(s^2))=32m`

Therefore, the stopping distance for the car is 32 meters.