Two cars started at the same position and were driven through a city using different streets. The red car traveled north 2 miles
1 answer:
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The impulse of a force is due to the change in the motion of an object
A. The persons speed after impact is approximately 59.38 mi/h
B. The expected speed is <u>29.89 mi/h</u> which is less than the findings
Reason:
Known parameters are;
The speed of the bus, v = 30 mi/h
The force with which the person was hit, F = 58,000 lbs
Mass of the bus, M = 40,000 lbs
Mass of the person, m = 150 lbs
Duration of the impact, Δt = 0.007 seconds
A. The speed of the person at the end of the impact, <em>v</em>, is given as follows;
The impulse of the force = F × Δt = m × Δv
For the person, we get;
58,000 lbf ≈ 1866094.816 lb·ft./s²
58,000 lbf × 0.007 s = 150 lbs × Δv
1,866,094.816 lb·ft./s²
Δv = v₂ - v₁
The initial speed of the person at the instant, can be as v₁ = 0
The final speed, v₂ = Δv - v₁
∴ v₂ ≈ 87.084 ft./s - 0 = 87.084 ft./s
≈ <u>87.084 ft./s</u>
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The speed of the person at the end of the impact, v₂ ≈ <u>59.38 mi/h</u>
B. Where the momentum is conserved, we have;
m₁·v₁ + m₂v₂ = (m₁ + m₂)·v
The expected speed of the person at the end of the impact is 29.89 mi/h, and therefore, <u>the findings does not agree with the expectation</u>
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Answer:
The speed of the plank relative to the ice is:
Explanation:
Here we can use momentum conservation. Do not forget it is relative to the ice.
(1)
Where:
- m(g) is the mass of the girl
- m(p) is the mass of the plank
- v(g) is the speed of the girl
- v(p) is the speed of the plank
Now, as we have relative velocities, we have:
(2)
v(g/b) is the speed of the girl relative to the plank
Solving the system of equations (1) and (2)
I hope it helps you!