Determine the stopping distance (in meters) if the brakes supply a force of 3500 newtons.

Question

A 2200 kg car is moving at a speed of 44.0m/s.

Answer 1

F= ma so a = F/m

Deceleration (a) is then 3500/2200 = 1.59091 however it is negative since the force is applied in the opposite direction as the travel. So a = - 1.59091

Using the kinematic equation:

v_f² = v_i² + 2ax we can solve for distance (x) as follows:

v_f² - v_i² = 2ax

(v_f² - v_i²)/(2a) = x

Plugging in the numbers (vf = 0, vi = 44, a = - 1.59091) we get:

x = (0² - 44²)/(2(- 1.59091))

x = -1936/(-3.1818)

x = 608.46 m

Since there are 2 significant figures, x = 610 m