Ask Question
15 July, 21:57

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?

+2
Answers (1)
  1. 15 July, 23:11
    0
    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.
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers