Motion NCERT Solution, study material, CBSE Notes

Motion NCERT Solution, study material, CBSE Notes

NCERT Solution: Motion

Using following data, draw time-displacement graph for a moving object:

Using following data, draw time-displacement graph for a moving object:

Use this graph to find average velocity for first 4 s, for next 4 s and for last 6 s.

Average velocity for first 4s.

(or as x remains the same from 4 to 8 seconds, velocity is zero)

An electron moving with a velocity of 5× 10⁴ ms^-1 enters into a uniform electric field and acquires a uniform acceleration of 10⁴ms^–2 in the direction of its initial motion. (i) Calculate the time in which the electron would acquire a velocity double of its initial velocity. (ii) How much distance the electron would cover in this time?

Given initial velocity, u = 5 � 10⁴ m/s

and acceleration, a = 10⁴ m s^�2

(i) final velocity = v = 2 u = 2 � 5 �10 ⁴ m s^�1

=10 � 10⁴ m s^�1

Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds.

Using the equation of motion

Distance travelled in the interval between 4th and 5th second

Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of

(Assume upward acceleration is -g and downward acceleration to be +g.

Science Sample paper 1

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