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 0.8 m/s2 for 4.2 seconds. It then continues at a constant speed for 12.7 seconds, before getting tired and slowing down with constant acceleration coming to rest 59 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. 1) How fast is the hare going 2.9 seconds after it starts? 2) How fast is the hare going 9.9 seconds after it starts? 3) How far does the hare travel before it begins to slow down? 4) What is the acceleration of the hare once it begins to slow down? Submit 5) What is the total time the hare is moving? 6) What is the acceleration of the tortoise?

Explanation / Answer

1) at t = 2.9 s

V_hare = vo + a*t

= 0 + 0.8*2.9

= 2.32 m/s

2) at t = 9.9 s

the hare is accelerated only for 4.2 s

so, V_hare = vo + a*t

= 0 + 0.8*4.2

= 3.36 m/s

3) distance travelled by hare before slwing down = 0.5*a*t1^2 + v*t2

= 0.5*0.8*4.2^2 + 3.36*12.7

= 49.73 m

4) let v1 = 3.36 m/s, v2 = 0

d = 59 - 49.73

= 9.27 m

so, apply, a = (v2^2 - v1^2)/(2*d)

= (0^2 - 3.36^2)/(2*9.27)

= -0.609 m/s^2

5) use, v2 = v1 + a*t3

==> t3 = (v2 - v1)/a

= ( 0 - 3.36)/(-0.609)

= 5.5 s

so, total time , T = t1 + t2 + t3

= 4.2 + 12.7 + 5.5

= 22.4 s

6) now use

L = 0.5*a*T^2

a = 2*L/T^2

= 2*59/22.4^2

= 0.235 m/s^2

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