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# Answer to Question #15225 in Mechanics | Relativity for fritz

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.
1
2012-09-25T11:07:42-0400
We know that

V = at ==&gt; t = V/a,

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

L = at&sup2;/2 ==&gt; a = 2L/t&sup2;,

where L is the stop distance. So,

a = 2L/t&sup2; = 2L/(V/a)&sup2; = 2L/(V&sup2;/a&sup2;) = 2a&sup2;L/V&sup2;

and

1 = 2aL/V&sup2; ==&gt; a = V&sup2;/(2L) = (100.0[km/h])&sup2;/(2*37.19[m]) = (100.0*1000/3600[m/s])&sup2;/(2*37.19[m]) &asymp; 10.37 m/s&sup2;.

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