means that a body is in motion, and its velocity is measured in meters per second. And, that velocity is increasing by two meters per second, every second.
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
Answer:
the car with the hay should slow to 16m/s if the bale of hay is dropped into it.
Answer:
A coefficient friction is a value that shows the relationship between two objects and the normal reaction between the objects that are involved.
Explanation:
We will use the formula / equation to determined the time.
Distance = ½ * (vi + vf) * t
48100 = ½ * (26.3 + 41.9) * t
t = 48100 ÷ 34.1 = 1410.557185 seconds
We will use the formula / equation to determined the acceleration.
vf = vi + a * t
41.9 = 26.3 + a * 1410.557185
a = (41.9 – 26.3) ÷ 1410.557185 = 0.011059459 m/s^2
We will use the formula / equation to determined the acceleration.
vf^2 = vi^2 + 2 * a * d
41.9^1 = 26.3^2 + 2 * a * 48100
a = (41.9^2 – 26.3^2) ÷ 96200 = 0. 011059459 m/s^2
Since both answers are the same, I believe the acceleration is correct.
