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AleksandrR [38]
3 years ago
15

Bus starts from rest if the acceleration of the bus is 0.5 MS squared what will be the velocity at the end of two minutes and wh

at distance will it cover during that time
Physics
1 answer:
Nutka1998 [239]3 years ago
4 0

Explanation:

Given that,

Initial speed of the bus, u = 0

Acceleration of the bus, a = 0.5 m/s²

Let v is the velocity at the end of 2 minutes. The change in velocity divided by time equals acceleration.

So,

a=\dfrac{v-u}{t}\\\\v=u+at\\\\v=0+0.5\times 120\\\\v=60\ m/s

Let d is the distance cover during that time. So,

v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{(60)^2}{2\times 0.5}\\\\d=3600\ m

So, the final speed is 60 m/s and the distance covered during that time is 3600 m.

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Answer:

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A ball is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the wa
vovangra [49]

Answer:

a) 48.5 ft/s

b) 36.5 ft

c) -80.3 ft/s

Explanation:

a)

The equation of motion of the ball is :

y(t) = -16.1 ft/s^2 * t^2 + Vo*t

Where Vo is the initial velocity

If y(5s) = - 160 ft:

-160 ft = -16.1 ft/s^2 * (5 s)^2 + Vo*(5s)

Solving for Vo

Vo  = (16.1*25- 160) ft / 5s = 48.5 ft/s

b)

To answer this question we must first know when the velocity became zero, at this time is when the ball was at its highest point.

v(t) = -32.2 ft/s^2 * t + Vo

t = Vo/32.2ft/s^2 = 1.5 s

And now, the highest point which the ball reached is given by:

y(1.5s) = -16.1 ft/s^2 * (1.5)^2 + Vo*(1.5s)

y(1.5s) = 36.52 ft

c)

We now need the time at which y(t') = -64 ft

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By means of the quadratic formula, we find that

t' = 4.00498 s ≈ 4 s

And the velocity at t = 4s is:

v(4s) = -32.2 ft/s^2 * 4s +48.5 ft/s = -80.3 ft/s

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3 years ago
An ant crawls 50 cm to the right, stops, and then crawls 30 cm to the left. Calculate the distance that the ant travels, and cal
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Answer and Explanation:

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Answer:

Good question to ask in physics, sir maam

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