6. A car is approaching a hill at 30 m/s when its engine is stopped at the bottom of the hill. The car moves up the hill with an acceleration of - 2 m/s/s. Determine the maximum distance the car travels up the hill.

225 m.

Explanation:

Applying the equation of motion.

v² = u²+2as Equation 1

Where v = final velocity of the car, u = initial velocity of the car, a = acceleration of the car, s = distance traveled by the car.

make s the subject if the equation.

s = (v²-u²)/2a  Equation 2

Given: v = 0 m/s, u = 30 m/s, a = -2 m/s²

Substitute into equation 2

s = (0²-30²)/(-2×2)

s = -900/-4

s = 225 m.

Hence the maximum distance the car travels up the hill =  225 m.

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000