# If a 2,400 lb. car is moving at 1 mile per hour, how much force is required to stop its motion? [closed]

Given that I stand behind or in front of a car and someone will move the car from within at approximately 1 M.P.H., how much force is needed to stop the force driven by the engine to put the car at rest?

Someone gave a random guess that it's a 1 lb. per pound of car force. In other words, exactly the car's weight in force multiplied by the totality of miles per hour. Sounded a little too simple for me.

How does this figure?

Example:

This is not any homework assignment; this is just me randomly wondering, because I've stood behind cars reversing slowly and no amount of force I've ever observed can stop a car (from humans) even going at the slowest rate possible. I don't know how to get a sufficient answer, so I can't offer much.

• Acceleration is a rate of change of velocity, so it should have units of length/(time*time). – Kyle Kanos Oct 7 '14 at 18:34
• Might as well consider acceleration to be movement/speed since change of velocity is of no importance here (we have no need to measure its rate of change but its actual speed in relation to its weight and how much force will stop it). I don't believe F = ma will provide a sufficient answer. – Ben Wa For Girls Oct 7 '14 at 18:36
• Sounds like OP is trying to ask 'If a 2,400 lb. car is travelling at 1 mile per hour, how much force is required to stop its motion?' OP, there's a huge difference between travelling at 1 mph and accelerating. – user121330 Oct 7 '14 at 18:38
• In short, you're asking an underdefined problem. That's why you're getting downvotes. – user121330 Oct 7 '14 at 18:50
• In order to relate a force to the velocity, a duration (i.e., length of time) needs to be defined. Are we talking 1s? 10s? An hour? Something else? Give us that, and then it can be computed. – Kyle Kanos Oct 7 '14 at 19:42

• The math behind this: Momentum of car (mass $m$, velocity $v$) = $m\cdot v$. You need force $F$ for time $\Delta t$ such that $F\Delta t = m\cdot \Delta v$. Then the rest of the above follows. – Floris Oct 7 '14 at 22:21