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 1.4 m/s2 for 4.9 seconds. It then continues at a constant speed for 11 seconds, before getting tired and slowing down with constant acceleration coming to rest 102 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 2.4 seconds after it starts?
2) How fast is the hare going 8.9 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?

1) Vf = 3.36 m/s

2) v = 6.86 m/s

3) s = 92.3 m

4) a = - 0.23 m/s²

Explanation:

1)

We use first equation of motion in this case:

Vf = Vi + at

where,

Vf = Final velocity = ?

Vi = Initial velocity = 0 m/s

a = acceleration = 1.4 m/s²

t = time = 2.4 s

Therefore,

Vf = 0 m/s + (1.4 m/s²)(2.4 s)

Vf = 3.36 m/s

2)

We again use first equation of motion but with t= 4.9 s now:

Vf = 0 m/s + (1.4 m/s²)(4.9 s)

Vf = 6.86 m/s

Now, this velocity remained constant for net 11 seconds. Hence, the velocity of hare after 8.9 s is:

v = 6.86 m/s

3)

First we use second equation of motion to find distance covered in accelerated motion:

s₁ = Vi t + (0.5)at²

s₁ = (0 m/s)(4.9 s) + (0.5)(1.4 m/s²)(4.9 s)²

s₁ = 16.8 m

Now, we calculate the distance covered in uniform motion:

s₂ = vt

s₂ = (6.86 m/s)(11 s)

s₂ = 75.5 m

Now, total distance covered before slowing down is given as:

s = s₁ + s₂ = 16.8 m + 75.5 m

s = 92.3 m

3)

Using third equation of motion for the decelerated motion:

2as = Vf² - Vi²

where,

a = deceleration = ?

s = distance covered = 102 m

Vf = Final Speed = 0 m/s

Vi = Initial Speed = 6.86 m/s

Therefore,

(2)(a)(102 m) = (0 m/s)² - (6.86 m/s)²

a = - (47.06 m²/s²)/(204 m)

a = - 0.23 m/s²