2)How fast is the hare going 11.8 seconds after it starts 3)How far does the har

Question

2)How fast is the hare going 11.8 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?

5)What is the total time the hare is moving?
6)What is the acceleration of the tortoise?

7)Below is some space to write notes on this problem

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/s2 for 4.7 seconds. It then continues at a constant speed for 11.6 seconds, before getting tired and slowing down with constant acceleration coming to rest 71 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.

Explanation / Answer

1)

a = acceleration of hare = 0.9 m/s2

t = time taken = 1.9 sec

Vi = initial velocity = 0 m/s

using the equation

Vf = Vi + at

Vf = 0 + 0.9 x 1.9

Vf = 1.71 m/s

2)

Velocity of hare after 4.7 sec can be given as

Vf = 0 + 4.7 x 0.9 = 4.23 m/s

Distance covered in 4.7 s can be given using the equation

d1 = Vi t + (0.5) a t2

d1 = 0 x 4.7 + (0.5) (0.9) (4.7)2

d1 = 9.941 m

distance travelled in remaining 11.8 - 4.7 = 7.1 s

d2 = 4.23 x 7.1 = 30.03 m

Total distance travelled = D = d1 + d2 = 9.941 + 30.03 = 39.97 m

3)

Velocity of hare after 4.7 sec can be given as

Vf = 0 + 4.7 x 0.9 = 4.23 m/s

Distance covered in 4.7 s can be given using the equation

d1 = Vi t + (0.5) a t2

d1 = 0 x 4.7 + (0.5) (0.9) (4.7)2

d1 = 9.941 m

distance travelled in remaining 11.6 s

d2 = 4.23 x 11.6 = 49.07 m

Total distance travelled = D = d1 + d2 = 9.941 + 49.07 = 59.01 m

4)

Distance Covered by hare while slowing down = 71 - distance covered before starting to slow down = 71 - 59.01 = 11.99 m

initial velocity = Vi = 4.23 m/s

final velocity = Vf = 0

using the equation

Vf2 = Vi2 + 2 a d

02 = 4.232 + 2 a (11.99)

a = - 0.75 m/s2

5)

time taken by hare to come to stop

t = Vf - Vi / a = 0 - 4.23 / (-0.75) = 5.64 sec

Total time hare was in motion = 4.7 + 11.6 + 5.64 = 21.94 sec

6)

d = distance covered = 71 m

time taken = t = 21.94 sec

Vi= initial velocity = 0

using the equation

d = Vi t + (0.5) a t2

71 = 0 (21.94) + (0.5) a (21.94)2

a = 0.295 m/s2

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