Motion CBSE Class 9 Physics Worksheets Set 09

Questions with Answers

Question 1. The figure shows the velocity-time graph of a body moving with initial velocity u and uniform acceleration a, reaching velocity v at time t. Using the graph find an expression connecting the distance s, u, a and t.

Answer:
The distance is the area under the graph, a trapezium with parallel sides u and v and width t: \( s = \tfrac12 (u + v)t \).
Since the slope of the graph is a, \( v = u + at \). Substituting:
\( s = \tfrac12 (u + u + at)t \), so \( s = ut + \tfrac12 at^2 \).
In simple words: The area of the trapezium is the rectangle ut plus the triangle with height at.

Teacher's Note:
1. The area under a v-t graph is the displacement.
2. The slope of the graph gives the acceleration.

Question 2. A stone is dropped from a tall tower. How much distance does it cover in the third second? (g = 9.8 m/s²)
Answer:
Distance in 3 s = \( \tfrac12 (9.8)(3^2) = 44.1 \) m. Distance in 2 s = \( \tfrac12 (9.8)(2^2) = 19.6 \) m.
Distance in the third second = 44.1 - 19.6 = 24.5 m.
In simple words: Subtract the distance covered in two seconds from that covered in three seconds.

Teacher's Note:
1. Alternatively use \( s_n = u + \dfrac{g}{2}(2n - 1) \).
2. The distances in successive seconds from rest are in the ratio 1 : 3 : 5.

Question 3. A body starts from a point and moves with a uniform velocity of 2 m/s for 8 seconds. It then reverses its direction and reaches the starting point again with uniform velocity, as shown in the displacement-time graph. Calculate the return velocity.

Answer:
Displacement at P after 8 s = 2 x 8 = 16 m. The body reaches the start at Q at t = 12 s.
Time taken to return = 12 - 8 = 4 s.
Return velocity = 16 / 4 = 4 m/s (directed back towards the start, so the slope of PQ is -4 m/s).
In simple words: It covers the same 16 m in half the time, so it goes twice as fast.

Teacher's Note:
1. The slope of a displacement-time graph is the velocity.
2. The graph rises for the first 8 s and falls for the last 4 s.

Question 4. A balloon moves up vertically from the ground at a uniform speed of 14.7 m/s. At the end of 6 seconds a bag is released from it. What is the time taken by the bag to reach the ground? (g = 9.8 m/s²)
Answer:
Height at release = 14.7 x 6 = 88.2 m, and the bag has an initial velocity of +14.7 m/s.
\( -88.2 = 14.7t - 4.9t^2 \Rightarrow t^2 - 3t - 18 = 0 \Rightarrow t = 6 \) s (rejecting -3).
Time taken = 6 s.
In simple words: The bag first goes up for 1.5 s and then falls down through the greater height.

Teacher's Note:
1. The bag starts with the velocity of the balloon.
2. Take upward as positive and the displacement as -88.2 m.

Question 5. A uniformly accelerating body covers a distance of 10 m in the 3rd second and 16 m in the 6th second. What are the initial velocity and the acceleration?
Answer:
\( s_n = u + \dfrac{a}{2}(2n - 1) \): \( u + 2.5a = 10 \) and \( u + 5.5a = 16 \).
Subtracting, 3a = 6, so a = 2 m/s². Then u = 10 - 5 = 5 m/s.
u = 5 m/s and a = 2 m/s².
In simple words: Two equations, two unknowns.

Teacher's Note:
1. The n-th second formula avoids subtracting two distances.
2. Check with the 6th second: 5 + 11 = 16 m.

Question 6. A body falling freely from rest covers 7/16 of the total height in the last second of its fall. Calculate the time of fall and the height from which it falls. (g = 9.8 m/s²)
Answer:
Let the time be t. Distance in the last second = \( \dfrac{g}{2}(2t - 1) = \dfrac{7}{16} \times \dfrac{g}{2}t^2 \).
\( 7t^2 - 32t + 16 = 0 \Rightarrow t = \dfrac{32 \pm 24}{14} \), so t = 4 s (4/7 s is rejected as the fall lasts longer than one second).
t = 4 s and height = \( \tfrac12 \times 9.8 \times 16 = 78.4 \) m.
In simple words: Express the last-second distance in terms of t and compare with 7/16 of the total.

Teacher's Note:
1. Reject any value of t below 1 s.
2. Check: 4.9 x 7 = 34.3 m = 7/16 of 78.4 m.

Question 7. A car starting from rest and moving with uniform acceleration reaches a speed of 90 km/h in 5 minutes. Calculate (i) its acceleration and (ii) the distance covered.
Answer:
90 km/h = 25 m/s and 5 min = 300 s.
(i) \( a = \dfrac{25 - 0}{300} = \dfrac{1}{12} \) m/s².
(ii) \( s = \tfrac12 (0 + 25)(300) = 3750 \) m. a = 1/12 m/s² and s = 3.75 km.
In simple words: Change the units, then use the equations of motion.

Teacher's Note:
1. Divide km/h by 3.6 to get m/s.
2. Distance = average velocity x time.

Question 8. A body starting from P travels 2 km in one hour towards east to a point Q and then another 1.5 km due north, also in one hour, to a point R. Calculate (a) its average speed and (b) its average velocity from P to R.
Answer:
(a) Average speed = total distance / total time = 3.5 km / 2 h = 1.75 km/h.
(b) Displacement PR = \( \sqrt{2^2 + 1.5^2} = 2.5 \) km, so the average velocity = 2.5 / 2 = 1.25 km/h (in the direction of PR, north of east).
Average speed = 1.75 km/h; average velocity = 1.25 km/h.
In simple words: Speed counts the path walked; velocity only the straight-line gap.

Teacher's Note:
1. Average velocity is always less than or equal to average speed.
2. The direction of PR makes an angle of tan⁻¹(0.75), about 37°, with east.

Question 9. A body moving from rest with a uniform acceleration of 0.5 m/s² for 25 seconds then changes its acceleration and moves with a uniform acceleration of -1.25 m/s² until it comes to rest. What is (a) the total distance covered and (b) the total time taken?
Answer:
After 25 s: v = 0.5 x 25 = 12.5 m/s and \( s_1 = \tfrac12 (0.5)(25)^2 = 156.25 \) m.
Retarding phase: time = 12.5 / 1.25 = 10 s and distance \( s_2 = \dfrac{12.5^2}{2 \times 1.25} = 62.5 \) m.
(a) Total distance = 156.25 + 62.5 = 218.75 m. (b) Total time = 25 + 10 = 35 s.
In simple words: Treat the two phases one after the other.

Teacher's Note:
1. The final velocity of the first phase is the initial velocity of the second.
2. Check with a velocity-time graph: area of the triangle = 1/2 x 35 x 12.5 = 218.75 m.

Question 10. A body undergoes three displacements: AB of 5 m towards the east, BC of 4 m in a direction 30° east of north and CD of 7 m towards the west. Calculate the resultant displacement.
Answer:
East components: 5 + 4 sin 30° - 7 = 5 + 2 - 7 = 0. North component: 4 cos 30° = \( 2\sqrt{3} \) m.
The resultant displacement AD = \( 2\sqrt{3} \approx 3.46 \) m in the direction north.
In simple words: Break each displacement into east and north parts and add them separately.

Teacher's Note:
1. Displacements are added as vectors.
2. The east components cancel completely.

Question 11. A body moving with uniform acceleration has an initial velocity of 2 m/s and a velocity of 10 m/s at the end of 5 seconds. Find the distance covered. Also find the velocity at the point midway in the journey and compare it with the average velocity.
Answer:
Distance = \( \dfrac{2 + 10}{2} \times 5 = 30 \) m.
Midway (by distance): \( v^2 = \dfrac{u^2 + v^2}{2} = \dfrac{4 + 100}{2} = 52 \Rightarrow v = 7.21 \) m/s.
Distance = 30 m; midway velocity = 7.21 m/s, which is greater than the average velocity of 6 m/s.
In simple words: The body spends more time in the first half of the distance, so its midway speed is above the average.

Teacher's Note:
1. Use v² = u² + 2as for the first half and for the whole.
2. The two velocities are different in uniformly accelerated motion.

Question 12. A piece of ice floats in a jar of water. What will happen to the level of the water when the ice melts completely? (Specific gravity of ice = 0.9)
Answer:
The floating ice displaces a mass of water equal to its own mass. When it melts, the melted ice becomes the same mass of water, which has the same volume as the water it had displaced.
The water level remains the same.
In simple words: Melted ice exactly fills the space the floating ice was pushing aside.

Teacher's Note:
1. Floating means the weight of the displaced water equals the weight of the ice.
2. If the ice were held under water the level would be different.