Schaum's 3000 Solved Problems in Physics · Problem 4.5 · Newton's laws

Stopping a car in thirty metres

No stopwatch in this one. Distance in, force out.

A 900-kg car is going 20 m/s along a level road. How large a constant retarding force is required to stop it in a distance of 30 m?as printed, page 51

Tap any symbol, here or in a formula, to see what it is and its value.

The picture
900 kg v = 20 m/s F = 6000 N, backward
time0.00 s
speed20.0 m/s
distance0.0 m
20 m/s0 3.0 s area so far = 0.0 m distance = area under the speed line whole triangle: ½ × 20 m/s × 3 s = 30 m
Feel the numbers
6000 Nbraking force

Two thirds of the car's own weight, pushing backward. Good tires on dry asphalt can do about this. On ice, not close.

0.68 g6.67 m/s²

A hard stop. You would be leaning into the seat belt, and anything loose on the seat would be on the floor.

3.0 stime to stop, not asked

The problem never mentions time, which is why v² = v₀² + 2ax is the formula. It falls out anyway: 20 divided by 6.67.

The walkthrough

Press play and it reads straight through, scrolling as it goes. Space pauses. ← → move a step. ▶ on a step plays from there.

Every symbol
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Narrated by the Listen voice. Timing comes from the narration itself, so the light lands on the part of the formula that is being said.
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