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# Category Archives: KINEMATICS

## Displacement – A Numerical Problem

A fly is sitting in the middle of a wall of a cubic room of side ’a’.If it flies and sit to one of opposite corner of the wall. Find its displacement?

If the fly sit on the same wall on the horizontal, (from A to B in the diagram) the displacement is “a
If the fly sits on the diagonally opposite corner on the same wall, (from A to C in the diagram) the displacement is “√2 a
If the fly sits on the corner on the longest diagonal of the room,  (from A to H in the diagram) the displacement is “√3 a

## Problem from Kinematics

Devika posted:

There are two particles A and B moving with same velocity u such that initially there velocity are perpendicular to each other and distance between them is L.the velocity of A is always directed towards B.find the distance between them after a long time.

Ans will be posted soon

## Ball falls to earth, and earth moves to ball?!!

“When the ball falls on the earth, the earth also moves up to meet it. But the motion of the earth is not noticeable. Explain why?”

The force exerted by the ball on earth is same as the force exerted by the earth on the ball. The acceleration of the ball towards earth is approximately 10 m/s2.

The acceleration of earth towards the ball = force/mass of earth ≈ 10-23 m/s2

When the earth moves towards the ball with this acceleration, the time required to travel the distance will be very large. If we assume a distance of 1m, and from the equation

$S=ut+\frac{1}{2}at^{2}$

we get t ≈ 1012 seconds whereas the ball reaches the earth in 0.4 seconds

[warning]In a nut shell, we can say that we cannot notice the motion of earth towards the ball due to the extremely low value of acceleration of earth towards the ball.[/warning]

## Another problem from kinematics

A boy jumps from rest, straight down from the top of a cliff. He falls halfway down to the water below in 0.800 s. How much time passes during his entire trip from the top down to the water? Ignore air resistance.

let the total height be h

So, for first case

S = h/2

a = g=10 m/s^2

u=0 m/s

t = 0.8 sec

using the relation $S=ut + \frac{1}{2}at^{2}$

h/2 = 0.5 x 10×0.8 ^2

h=6.4 m

In second case (considering the full motion)

S=h=6.4m

t=?

a=g=10m/s^2

u=o m/s

using the relation $S=ut + \frac{1}{2}at^{2}$
6.4 = 0.5 x 10 x t^2

or

t = 2 x 6.4/10 = 1.28 s

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