A car traveling at 40 ft/sec decelerates at a constant 5 feet per second per second. how many feet does the car travel before coming to a complete stop?

Question

Jeramiah Mills

Physics

6 October, 11:37

A car traveling at 40 ft/sec decelerates at a constant 5 feet per second per second. how many feet does the car travel before coming to a complete stop?

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Jabari Herman · Answer 1

This problem is solved by the equation of motion: x = x0 + v0*t + 1/2*a*t^2, Here x0 = 0, v0 = 40ft/sec and a = - 5 ft/s^2, we need to solve for t: v = v0 + a*t, solve how long does it take to stop: 0 = v0 + a*t - - > a*t = - v0 - - > t = - v0/a - - > 40/5 = 8 seconds to stop. In this time, the car travels x = 0 + 40*8 + 0.5*-5*8^2 ft ~ 160 ft. Answer: The car travels 160 ft.