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 4.4 seconds. It then continues at a constant speed for 13.5 seconds, before getting tired and slowing down with constant acceleration coming to rest 61 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 2.2 seconds after it starts?   

How fast is the hare going 15.1 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?

the last 3 are most important

Answer 1

vf = vi + a t

vi = 0 and a = 0.8 m/s^2

v = 0 + (0.8 x 2.2) = 1.76 m/s ........Ans
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at t = 15.1 sec,hare is moving with constant velocity.

v = 0 + (0.8 x 4.4) = 3.52 m/s .......Ans

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distance travelled in first 4.4 sec ,

d1= (0 x 4.4) + (0.8 x 4.4^2 / 2) = 7.74 m

after that distance travelled in 13.5 sec,

d2 = (13.5 x 3.52) = 47.5 m

total distance travelled = d1 + d2 = 55.3 m .....Ans

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distance it travels while slowing down = 61 - 55.3 = 5.7 m

for this displacement,. v0 = 3.52 m/s

vf = 0

vf^2 - vi^2 = 2 a d

0^2 - 3.52^2 = 2 (a) (5.7)

a = - 1.09 m/s^2 .....Ans
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time taken to slow down, 0 = 3.52 - 1.09t1

t1 = 3.24 sec

t = 5.5 + 13.4 + 3.24 = 22.1 sec .......Ans

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d = 61 m

t = 22.1 sec

d = v0 t + a t^2 / 2

61 = 0 + a(22.1^2)/2

a = 0.25 m/s^2 ......Ans