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Here you get some question answers for practicing motion chapter
Typology: Exams
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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
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