A cheetah can accelerate at 4.5 m/s from rest to a speed of 30.0 m/s. Calculatethe distance
1 answer:
Answer:
d= 100m
Explanation:
Cheetah kinematic
The cheetah moves with uniformly accelerated movement, and the formulas that describe this movement are:
d= v₀*t + (1/2)*a*t² Formula (1)
vf²=v₀²+2*a*d Formula (2)
vf=v₀+a*t Formula (3)
Where:
d:distance in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
t: time in seconds (s)
Known Data
v₀ =0
a = 4.5 m/s²
vf= 30 m/s.
Problem development
We apply the formula (2) that has known data to calculate the distance :
vf²=v₀²+2*a*d
(30)²= 0 + 2* 4.5* d
d= 100m
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Answer:
Δθ₁ = 172.5 rev
Δθ₁h = 43.1 rev
Δθ₂ = 920 rev
Δθ₂h = 690 rev
Explanation:
- Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:
- Now, we need first to find the value of the angular acceleration, that we can get from the following expression:
- Since the machine starts from rest, ω₀ = 0.
- We know the value of ωf₁ (the operating speed) in rev/min.
- Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:
- Replacing by the givens in (2):
- Solving for α:
- Replacing (5) and Δt in (1), we get:
- in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:
- In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:
- First of all, we need to find the value of the angular acceleration during the second period.
- We can use again (2) replacing by the givens:
- ωf =0 (the machine finally comes to an stop)
- ω₀ = ωf₁ = 57.5 rev/sec
- Δt = 32 s
- Solving for α in (9), we get:
- Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:
- In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows: