Answer:
A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes
Explanation:
Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
where,
Vf = Final Velocity of Car = 0 mi/h
Vi = Initial Velocity of Car = 50 mi/h
a = deceleration of car
s = distance covered
Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.
So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:
s = vt
FOR SOBER DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 0.33 s
s = s₁
Therefore,
s₁ = (73.33 ft/s)(0.33 s)
s₁ = 24.2 ft
FOR DRUNK DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 1 s
s = s₂
Therefore,
s₂ = (73.33 ft/s)(1 s)
s₂ = 73.33 ft
Now, the distance traveled by drunk driver's car further than sober driver's car is given by:
ΔS = s₂ - s₁
ΔS = 73.33 ft - 24.2 ft
<u>ΔS = 49.13 ft</u>
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Answer:
I = I₀ + M(L/2)²
Explanation:
Given that the moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is I₀.
The parallel axis theorem for moment of inertia states that the moment of inertia of a body about an axis passing through the centre of mass is equal to the sum of the moment of inertia of the body about an axis passing through the centre of mass and the product of mass and the square of the distance between the two axes.
The moment of inertia of the body about an axis passing through the centre of mass is given to be I₀
The distance between the two axes is L/2 (total length of the rod divided by 2
From the parallel axis theorem we have
I = I₀ + M(L/2)²
In a circle or circular motion
