UNIT 2:

Kinematics
Numerical Problems...

Numerical No: 2.1,2.2,2.3,2.4,2.5,2.6,2.7,2.8,2.9,2.10

NUMERICAL NO: 2.1

• A train moves with a uniform velocity of 36 kmh^-1 for 10 s. find the distance traveled by it.

SOLUTION...

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NUMERICAL NO: 2.2

A train starts from rest. It moves through 1 km in 100 s with uniform acceleration. what will be its speed at the end of 100 s?

SOLUTION...

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NUMERICAL NO: 2.3

A car has a velocity of 10 ms^-1 for half minute. Find the distance traveled during this time and the final velocity of the car.

SOLUTION...

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NUMERICAL NO: 2.4

A tennis ball is hit vertically upward with a velocity of 30 ms^-1. It takes 3 s to reach the highest point. Calculate the maximum height reached by the ball. How long it will take to reach the ground?

SOLUTION...

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NUMERICAL NO: 2.5

A car moves with uniform velocity of 40 ms^-1 for 5 s. It comes to rest in the next 10 s with uniform deceleration. find:

• deceleration
• total distance traveled by the car

SOLUTION...

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NUMERICAL NO: 2.6

A train starts from rest with an acceleration of 0.5 ms^-2. Find its speed in kmh^-1, when it has moved through 100 m.

SOLUTION...

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NUMERICAL NO: 2.7

A train starting from rest, accelerates uniformly and attains a velocity of 48 kmh^-1 in 2 minutes. it travels at this speed for 5 minutes. finally, it moves with uniform retardation and it stopped after 3 minutes. Find the total distance traveled by the train.

SOLUTION...

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NUMERICAL NO: 2.8

A cricket ball vertically upwards and returns to ground 6 s later. Calculate;

• Maximum height reached by the ball.
• Initial velocity of the ball.

SOLUTION...

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NUMERICAL NO: 2.9

When brakes are applied, the speed of train decreases from 96 kmh^-1 to 48 kmh^-1 in 800 m. How much long will it take before coming to rest? (Assuming the retardation to be constant.)

SOLUTION...

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NUMERICAL NO: 2.10

In the above problem (Numerical No. 2.9), find the time taken by the train to stop after the application of brakes.

SOLUTION...

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