According to recent test data, an automobile travels 0.250 mi in 19.9 s, starting from rest. The same car, when braking from 60.0 mi/h on dry pavement, stops in 146 ft. Assume constant acceleration in each part of the motion, but not necessarily the same acceleration when slowing down as when speeding up.

Question

If its acceleration is constant, how fast (in mi/h) should this car be traveling after 0.250 mi of acceleration? The actual measured speed is 70.0 mi/h what does this tell you about the motion?

Answer 1

Ab = V²/2x = -88²/(2*146) = -26.52 ft/sec²

Aa = 2x/t² = 2(5280/4)/19.9² = 6.666 ft/sec²

V = √[2ax] = √[2*6.666*5280/4] = 132.658 ft/sec = 90.4 mph

t = √[2x/a] = √[2*5280/(4*26.52)] = 9.98 sec