A sprinter, in the 100m dash, accelerates from rest to a top speedwith a (consta

Question

A sprinter, in the 100m dash, accelerates from rest to a top speedwith a (constant) acceleration of 2.40 m/s2 andmaintains the top speed to the end of the dash. The total timetaken in the dash is 13.1s.
(a) What was the time elapsed during the acceleration phase of thesprinter's motion?
(b) What distance did the sprinter cover during the accelerationphase of his motion?
(c) What was the top speed reached by the sprinter?

Explanation / Answer

Let T be the time during the acceleration phase => the top velocity : V = a*T = 2.4T the distance on the the acceleration phase : d = (1/2)aT^2 = 1.2T^2 it take the time t =13.1 - T to complete the dash V*t = 100- d => 2.4T*(13.1-T) = 100 - 1.2T^2 => 1.2T^2 -31.44T + 100 =0 => T = 3.7 s   (another rootis larger than 13.1 s) b) =>distance : d = 1.2T^2 = 16.428m c) top speed : V = 2.4*T = 8.88m/s

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