Question:
A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?
Answer:
Time for the race will be t = 9.26 s
Explanation:
Given data:
As the sprinter starts the race so initial velocity = v₁ = 0
Distance = s₁ = 20 m
Acceleration = a = 4.20 ms⁻²
Distance = s₂ = 100 m
We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.
Using 3rd equation of motion
(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)
v₂ = 12.96 ms⁻¹
Time for 20 m distance = t₁ = (v₂ - v ₁)/a
t₁ = 12.96/4.2 = 3.09 s
He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be
Time for 100 m distance = t₂ = s₂/v₂
t₂ = 80/12.96 = 6.17 s
Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s
T = 9.26 s
The correct answer would have to be the pitch would increase
Answer:
497.6 N
Explanation:
From the question,
The net force on the skydiver = weight of the skydiver- the resistive force of air
F' = W-F...................... Equation 1
Where W = weight of the skydiver, F = resistive force of air.
But,
W = mg................ Equation 2
Where m = mass of the skydiver, g = acceleration due to gravity.
Substitute equation 2 into equation 1
F' = mg-F............ Equation 3
Given: m = 87 kg, F = 355 N, g = 9.8 m/s²
Substitute these values into equation 3
F' = 87(9.8)-355
F' = 852.6-355
F' = 487.6 N
