Answer:
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)
<u>Vf = 3.36 m/s</u>
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:
<u>v = 6.86 m/s</u>
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
<u>s = 92.3 m</u>
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)
<u>a = - 0.23 m/s²</u>
