1Revise, Reflect, Refine16 questions
Q.1My father went to a shop from home which is located at a distance of 250 m on a straight road. On reaching there, he discovered that he forgot to carry a cloth bag. He came home to take it, went to the shop again, bought provisions and came back home. How much was the total distance travelled by him? What was his displacement from home?v
Answer:Total distance = 1000 m; displacement = 0 m.
Q.2A student runs from the ground floor to the fourth floor of a school building to collect a book and then comes down to their classroom on the second floor. m, find: (i) the total vertical distance travelled, and (ii) their displacement from the starting point.v
Answer:(i) 18 m (ii) 6 m upward.
Q.3A girl is riding her scooter and finds that its speedometer reading is constant. Is it possible for her scooter to be accelerating and if so, how?v
Answer:Yes. If speed is constant but direction changes, velocity changes, so the scooter is accelerating.
Q.4m s–1 in 6 s. Find the average acceleration and the distance travelled in these 6 s.v
Answer:Acceleration = 4 m s⁻²; distance = 72 m.
Q.5m s–1 and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.v
Answer:Acceleration = –4 m s⁻²; time = 7 s.
Q.6Fig. 4.27 shows a position-time graph of two objects A and B that are moving along the parallel tracks in the same direction. Do objects A and B ever have equal velocity? Justify your answer. A (m) (m) B A B 5 0 10 Time (s) Time (s) Fig. 4.27 Fig. 4.28v
Answer:No. Equal velocity would require equal slopes on the position-time graph.
Q.7A graph in Fig. 4.28 shows the change in position with time for two objects to 10 seconds. Choose the correct option(s). (i) s time interval is equal since they have the same initial and final positions. (ii) s time interval are equal since both cover equal distance in equal time. (iii) s time interval is lower than that of seconds. (iv) s time interval is greater than that of B since B’s speed is lower than A’s in some segments.v
Q.8km h–1 notices a road sign गति सीमा km h–1 (Fig. 4.29) for trucks. He slows down to 36 km h–1 in 36 s. What was the distance travelled by him during this Fig. 4.29 time? Assume the acceleration to be constant while slowing down.v
Q.9m s–1 in 5 seconds. m s–1 for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.v
Q.10km h–1 when the driver sees an obstacle 30 m ahead. .5 seconds to react before pressing the brake. Once the brake is applied, the velocity of the bus reduces with constant acceleration of 2.5 m s–v
Answer:The bus travels 25 m before stopping, so yes, it stops before the obstacle.
Q.11A student said, “The Earth moves around the Sun”. In this context, discuss whether an object kept on the Earth can be considered to be at rest.v
Answer:Rest and motion depend on the frame of reference. The object can be at rest relative to Earth or a room, while it is moving relative to the Sun along with Earth.
Q.12s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas (in different colours) representing the displacement of the cyclist (i) while cyclist is moving with constant velocity. (ii) when the velocity of cyclist is decreasing. s time interval. 6 7.5 h–1) s–1) 5 (Km (m 4 5.0 3 Velocity 2.5 1 0 20 40 60 80 100 120 0 2 4 6 Time (s) Time (h) Fig. 4.30 Fig. 4.31v
Answer:Displacement is the area under the velocity-time graph. From Fig. 4.30, total displacement is 320 m and average acceleration over 120 s is 1/60 m s⁻². Shade the constant-velocity rectangular region and the decreasing-velocity triangular/trapezium region.
Q.13A girl is preparing for her first marathon by running on a straight road. She uses a smartwatch to calculate her running speed at different intervals. The graph (Fig. 4.31) depicts her velocity versus time. Estimate the distance she ran based on the graph.v
Solutiondistance = area under the velocity-time graph = sum of the areas of the triangular (accelerating) and rectangular (constant-speed) regions.
Answer:The distance covered equals the area enclosed between the velocity–time graph and the time axis. Split that area into simple shapes (rectangles and triangles), work out each area using the velocity (in m/s) and time (in s) read off the axes, and add them. For a runner who speeds up uniformly from rest to a steady velocity and then keeps it constant, for example reaching 5 m/s in 20 s and then running at 5 m/s for a further 40 s, the distance = area of the triangle (½ × 20 × 5 = 50 m) + area of the rectangle (5 × 40 = 200 m) = 250 m. Apply the same area method to the values shown on Fig. 4.31.
Q.14On entering a state highway, a car continues to move with a constant velocity of 6 m s–1 for 2 minutes and then accelerates with a constant acceleration 1 m s–2 for 6 seconds. Find the displacement of the car on the state highway in the 2 min 6 s time interval by drawing a velocity-time graph for its motion.v
Q.15Two cars A and B start moving with a constant acceleration from rest, in a straight line. m s–1 in 5 s. Car B attains a velocity of 3 m s–1 in 10 s. Plot the velocity-time graphs for both the cars in the same graph. Using the graph, calculate the displacement in the two time intervals mentioned (Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of time to plot the graph).v
Answer:Car A displacement = 12.5 m; Car B displacement = 15 m.
Q.16:30 PM at home. Consider the tip of the minute’s hand of the wall clock. During the given time interval, what is its: (i) distance travelled, (ii) displacement, (iii) speed, and (iv) velocity. The length of the minute’s hand is 7 cm (Fig. 4.32). Fig. 4.32v
Answer:(i) 66 cm (ii) 14 cm (iii) 11/900 cm s⁻¹ (iv) 7/2700 cm s⁻¹.