A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?

The displacement of the car after 6s is 43.2 m

Explanation:

Given;

velocity of the car, v = 12 m/s

acceleration of the car, a = -1.6 m/s² (backward acceleration)

time of motion, t = 6 s

The displacement of the car after 6s is given by the following kinematic equation;

d = ut + ¹/₂at²

d = (12 x 6) + ¹/₂(-1.6)(6)²

d = 72 - 28.8

d = 43.2 m

Therefore, the displacement of the car after 6s is 43.2 m

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.