Question #15225

The brochure advertising a sports car states that the car can be moving at 100.0 km/h, and stop in 37.19 meters. What is its average acceleration during a stop from that velocity? Express your answer in m/s2. Consider the car's initial velocity to be a positive quantity.

Expert's answer

We know that

V = at ==> t = V/a,

where V is velocity, a is acceleration and t is the time of acceleration. Also we know that

L = at²/2 ==> a = 2L/t²,

where L is the stop distance. So,

a = 2L/t² = 2L/(V/a)² = 2L/(V²/a²) = 2a²L/V²

and

1 = 2aL/V² ==> a = V²/(2L) = (100.0[km/h])²/(2*37.19[m]) = (100.0*1000/3600[m/s])²/(2*37.19[m]) ≈ 10.37 m/s².

V = at ==> t = V/a,

where V is velocity, a is acceleration and t is the time of acceleration. Also we know that

L = at²/2 ==> a = 2L/t²,

where L is the stop distance. So,

a = 2L/t² = 2L/(V/a)² = 2L/(V²/a²) = 2a²L/V²

and

1 = 2aL/V² ==> a = V²/(2L) = (100.0[km/h])²/(2*37.19[m]) = (100.0*1000/3600[m/s])²/(2*37.19[m]) ≈ 10.37 m/s².

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