NCERT Solutions for Chapter: Motion, Exercise 3: Long Answer Questions

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NCERT Science Solutions for Exercise - NCERT Solutions for Chapter: Motion, Exercise 3: Long Answer Questions

Attempt the practice questions on Chapter 8: Motion, Exercise 3: Long Answer Questions with hints and solutions to strengthen your understanding. NCERT Exemplar Science - Class 9 solutions are prepared by Experienced Embibe Experts.

Questions from NCERT Solutions for Chapter: Motion, Exercise 3: Long Answer Questions with Hints & Solutions

HARD
9th CBSE
IMPORTANT

An object is dropped from rest at a height of 150metre and simultaneously another object is dropped from rest at a height 100metre. What is the difference in their heights after 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time? 

HARD
9th CBSE
IMPORTANT

An object starting from rest travels 20 m in first 2 s and 160 m in next 4 s. What will be the velocity after 7 s from the start?

HARD
9th CBSE
IMPORTANT

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

Time (s) 0 2 4 6 8 10 12 14 16
Displacement (m) 0 2 4 4 4 6 4 2 0

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

HARD
9th CBSE
IMPORTANT

An electron moving with a velocity of 5×104 m s-1 enters into a uniform electric field and acquires a uniform acceleration of 104 m s-2 in the direction of its initial motion. Calculate the time in which the electron would acquire a velocity double of its initial velocity.

MEDIUM
9th CBSE
IMPORTANT

An electron moving with a velocity of 5×104 m s-1 enters into a uniform electric field and acquires a uniform acceleration of 104 m s-2 in the direction of its initial motion. How much distance the electron would cover in this time?

HARD
9th CBSE
IMPORTANT

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

HARD
9th CBSE
IMPORTANT

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 u12:u22 (Assume upward acceleration is -g and downward acceleration to be +g).