1- How fast is the hare going 3.9 seconds after it starts? 2-How fast is the har

Question

1- How fast is the hare going 3.9 seconds after it starts?

2-How fast is the hare going 10.3 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?

Tortoise and Hare w/ Solution 23 45 67 2 345 67 A tortoise and hare start from rest and ave a ace. As the race begins, both accelerate forward. The hare accelerates uniformly at a rate of 1 m/s2 for 4.9 seconds. It then continues at a constant speed for 12.6 seconds, before getting tired and slowing down with constant acceleration coming to rest 82 meters from where it started. The tortoise for the entire distance, fially catching the hare just as the hare comes to a stop.

Explanation / Answer

acceleration of ahre, ah = 1 m/s^2 for t = 4.9s
distance to constant acceleration, d = 82m

Part A:
Still in the initial acceleration phase
v = at
v = 1(3.3)
v = 3.3 m/s

Part B:

This is during the constant speed section
v = 1*(4.9)
v = 4.9 m/s

Part C:
this will be the sum of the distance covered under initial acceleration plus the distance covered by constant speed
From second equation of motion,
di = 0.5*a*t^2 + vt
di = 0.5*a*t1^2 + a*t1*t2
di = 0.5*1*4.9^2 + 1 * 4.9 * 12.6
di = 73.745 m

Part D:
Acceleration must bring speed to zero and cover the remaining distance in the same time , from third eqn of motion,

vf^2 = vi^2 + 2a(d - di)
0 = 4.9^2 + 2a(82 - 73.745)
a = -1.454 m/s^2
-ve sign shows deacceleration

Part E:
time for decelleration
v = at
t = v/a
t = 4.9 / 1.454
t = 3.37 s

Therefore net time for journey,
tn = 4.9 + 12.6 + 3.37
tn = 20.87 s

Part F:
From second Equation of motion
d = 0.5at^2
a = 2d/t^2
a = 2(82) / (20.87)^2
a = 0.377 m/s^2

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