A bus starts from rest accelerating uniformly with 4m/s2 at t = 10s a stone is dropped out of a window of the bus 2m high what are the

1. magnitude of velovity

2.acceleraton of the stone at 10.2s

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Solution

Let's deal with part b) first because it is the easiest: t = 10.2 s is after the stone was dropped out the window, which means that it is no longer accelerating horizontally. Its only acceleration will be the downwards acceleration due to gravity (10 m/s²).

Now, for part a)
The horizontal component of velocity will be equal to the velocity of the bus as it is released (because after this point the horizontal force is removed, keeping the velocity constant.

So, to find the speed of the bus at time t= 10 s:
t = time - 10 s
a = acceleration = 4 m/s²
u = initial velocity = 0 m/s
v = final velocity = ?
v = u + at
= 0 + 410 = 40 m/s (This is absurd by the way - 40 m/s = 144 km/h)

Vertical component of the stone's velocity? We have the following information:
t = time = 0.2 s (It's this long since the stone was released
g = acceleration due to gravity = 10 m/s²
u = initial velocity = 0 m/s
v = final velocity = ?

v = u + gt
= 0 + 100.2
= 2 m/s.

So, magnitude = √(40² + 2²)
= 40.0 m/s.

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