What average force is required to stop a 950-kg car in 8.0 s if the car is traveling at 95 km/h?
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This is Giancoli Answers with Mr. Dychko. Will start this question by writing down the stuff that we know and doing a little unit conversion here for the velocity. So, the mass of the car is 950 kilograms, it's going to be stopping within 8 seconds, and it's initially traveling at 95 kilometers an hour. And we'll multiply that by one hour for every 3600 seconds, and leaving us with seconds on the bottom and the hours cancel, and then times by 1,000 meters for every kilometer, and the kilometers cancel leaving us with meters in the top and we get 25.389 m/s. This is also the same as dividing by 3.6. And the final speed of the car is zero. Here's a free body diagram of the car, sort of you have neglected the vertical direction. But horizontally speaking we have the stopping force is going to be equal to the net force because it's the only force exerted on the car horizontally. And the car initially will say is going to the right. So, this makes the stopping force equal to ma since it's the net force. And so, we have to find acceleration. And we have this handy formula here to help us with that. Final velocity is zero, and we'll subtract V initial from both sides, and then divide both sides by t. And we get acceleration is initial velocity divided by time with a negative in front. And negative just means the acceleration will be in the opposite direction to the initial velocity. So, it's negative 26.389 m/s divided by 8 seconds which is negative 3.2986 m/s squared. So, we'll take that acceleration multiplied by the mass of 950 kilograms to get the stopping force with two significant figures is 3100 Newtons.
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