motion chapter physics, Exams of Physics

Here you get some question answers for practicing motion chapter

Typology: Exams

2025/2026

Available from 06/22/2026

sameer-mattu
sameer-mattu 🇮🇳

8 documents

1 / 3

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Motion – Easy Step-by-Step Solutions
TWO MARK QUESTIONS
1. Two differences between instantaneous speed and instantaneous velocity
Instantaneous speed is the speed at a particular instant and is a scalar quantity.
Instantaneous velocity is the velocity at a particular instant and is a vector quantity.
2. Position-time graph for constant speed
The graph is a straight line with constant slope.
3. Position-time graph for positive direction with negative acceleration
The graph is a curve with decreasing slope.
4. Average velocity and uniform velocity
Average velocity = Total displacement / Total time.
Uniform velocity means equal displacement in equal intervals of time in the same direction.
5. Velocity after 20 s
Given:
u = 5 m/s, a = 1 m/s², t = 20 s
v = u + at
= 5 + (1 × 20)
= 25 m/s.
6. Velocity-time graph explanation
(i) Ball thrown upward: velocity decreases to zero and becomes negative while returning.
(ii) Ball dropped and bouncing: velocity increases downward and changes direction after collision.
7. Ball thrown downward
Initial speed = 20 m/s.
After collision speed becomes 18 m/s.
Using:
v² = u² - 2gh
0 = 18² - 2 × 9.8 × h
h = 324 / 19.6
h = 16.53 m.
THREE MARK QUESTIONS
1. Derive v = u + at
a = (v - u)/t
at = v - u
v = u + at.
2. Derive S = ut + 1/2at²
Average velocity = (u + v)/2
S = [(u + v)/2] × t
Using v = u + at,
S = ut + 1/2at².
3. Derive v² - u² = 2as
Using:
v = u + at and S = [(u+v)/2]t
After substitution:
v² - u² = 2as.
pf3

Partial preview of the text

Download motion chapter physics and more Exams Physics in PDF only on Docsity!

Motion – Easy Step-by-Step Solutions

TWO MARK QUESTIONS

1. Two differences between instantaneous speed and instantaneous velocity Instantaneous speed is the speed at a particular instant and is a scalar quantity. Instantaneous velocity is the velocity at a particular instant and is a vector quantity. 2. Position-time graph for constant speed The graph is a straight line with constant slope. 3. Position-time graph for positive direction with negative acceleration The graph is a curve with decreasing slope. 4. Average velocity and uniform velocity Average velocity = Total displacement / Total time. Uniform velocity means equal displacement in equal intervals of time in the same direction. 5. Velocity after 20 s Given: u = 5 m/s, a = 1 m/s², t = 20 s

v = u + at = 5 + (1 × 20) = 25 m/s.

6. Velocity-time graph explanation (i) Ball thrown upward: velocity decreases to zero and becomes negative while returning. (ii) Ball dropped and bouncing: velocity increases downward and changes direction after collision. 7. Ball thrown downward Initial speed = 20 m/s. After collision speed becomes 18 m/s.

Using: v² = u² - 2gh 0 = 18² - 2 × 9.8 × h h = 324 / 19. h = 16.53 m.

THREE MARK QUESTIONS

1. Derive v = u + at a = (v - u)/t at = v - u v = u + at. 2. Derive S = ut + 1/2at² Average velocity = (u + v)/ S = [(u + v)/2] × t Using v = u + at, S = ut + 1/2at². 3. Derive v² - u² = 2as Using: v = u + at and S = [(u+v)/2]t After substitution: v² - u² = 2as.

4. Applications of velocity-time graph

  • To find acceleration
  • To find displacement
  • To study motion of body 5. Car stopping problem u = 126 km/h = 35 m/s v = 0, s = 200 m

Using: v² = u² + 2as 0 = 35² + 2 × a × 200 a = -3.06 m/s².

Retardation = 3.06 m/s².

6. Maximum height of stone u = 20 m/s, g = 10 m/s²

Using: v² = u² - 2gh 0 = 400 - 20h h = 20 m.

FIVE MARK QUESTION

1. Three equations of motion (i) v = u + at (ii) S = ut + 1/2at² (iii) v² = u² + 2as

NUMERICALS

1. Average velocity Total distance = 3d Time taken = d/20 + d/40 + d/ = 11d/

Average velocity = 3d ÷ (11d/120) = 360/ = 32.72 m/s.

2. Ball thrown upward with 29.4 m/s (i) Maximum height: 0 = 29.4² - 2 × 9.8 × h h = 44.1 m.

(ii) Time to highest point: 0 = 29.4 - 9.8t t = 3 s.

3. Body falling from tower Using formula for nth second motion, Height of tower = 125 m. 4. Body projected upward from tower u = 15 m/s, t = 5 s