A tortoise and hare start from rest and have a race. As the race begins, both ac

Question

A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly at a rate of 0.9 m/s^2 for 4.6 seconds. It then continues at a constant speed for 11.3 seconds, before getting tired and slowing down with constant acceleration coming to rest 67 meters from where it started. The tortoise accelerates uniformly for the entire distance, finally catching the hare just as the hare comes to a stop. How fast is the hare going 3.7 seconds after it starts? How fast is the hare going 10.2 seconds after it starts? How far does the hare travel before it begins to slow down? What is the acceleration of the hare once it begins to slow down? What is the total time the hare is moving? What is the acceleration of the tortoise?

Explanation / Answer

1) As for first 4.6 seconds the hare accelerates uniformly at a rate of 0.9 m/s^2 from rest, so the velocity at t = 3.7 sec, v = 3.7*0.9 = 3.33 m/s
2) At t= 4.6 sec the hare has a velocity v = 4.6*0.9 = 4.14 m/s and continues will the same velocity till t = 11.3 sec , so the velocity at t = 10.2 sec will be 4.14 m/s
3) Distance travelled by here in first 4.6 sec, d1 = 0.5*a*t^2 = 0.5*0.9*4.6^2 = 9.522 m and after this distance travelled in 11.3 sec is d2 = 11.3*4.14 =46.782 m so total distance travelled = 56.304 m
4) The distance travelled by hare while decelerating, d = 67-56.304 = 8.7 m
using v^2 = u^2+2*a*d, we get a = -0.985 m/s^2
5) Using v=u+a*t, we get t = 4.2 sec, so total travelling time = 4.2+11.3 = 15.5 sec.
6) using d = 0.5*a*t^2, where d = 67m, t = 15.5 sec, so we get a = 0.558 m/s^2.

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