Motion And Time

📘 Study MCQs

Q1. The formula to calculate time taken when distance and speed are known is:
A. Time taken = Distance × Speed
B. Time taken = Distance / Speed
C. Time taken = Speed / Distance
D. Time taken = Distance + Speed
B. Time taken = Distance / Speed
Speed is defined as distance divided by time. If you rearrange this formula, time taken equals distance divided by speed. For example, if a car has to cover 100 kilometre at a speed of 50 kilometre per hour, the time taken will be 100/50 = 2 hour.

Q2. An odometer in a vehicle measures:
A. Instantaneous speed
B. Total distance travelled
C. Time taken
D. Fuel remaining
B. Total distance travelled
An odometer is a device fitted in the dashboard of a vehicle. It records the total distance the vehicle has travelled since it was manufactured, usually in kilometre. It does not measure speed directly; a speedometer measures speed.

Q3. A car travels at a speed of 40 kilometre per hour. How much time will it take to cover 120 kilometre?
A. 2 hour
B. 3 hour
C. 4 hour
D. 5 hour
B. 3 hour
Using the formula time taken = distance / speed. Here distance = 120 km, speed = 40 km/h. So time = 120/40 = 3 hour. This means the car will need 3 hour to complete the journey.

Q4. If a cyclist covers 60 kilometre in 3 hour, the speed is:
A. 20 km/h
B. 30 km/h
C. 180 km/h
D. 63 km/h
A. 20 km/h
Speed = distance / time = 60 km / 3 h = 20 km/h. This is uniform speed if the cyclist does not change pace. For non-uniform motion, this would be the average speed.

Q5. A bar graph is used to represent data:
A. In the form of pictures
B. In the form of rectangles of equal width
C. In the form of a circle
D. In the form of a line
B. In the form of rectangles of equal width
A bar graph uses rectangular bars of equal width but different heights to show comparisons among different categories. The height of each bar represents the value of the data, such as the distance covered by different vehicles in one hour.

Q6. In a distance-time graph, time is usually plotted on which axis?
A. X-axis (horizontal axis)
B. Y-axis (vertical axis)
C. Z-axis
D. Both axes
A. X-axis (horizontal axis)
By convention, in a distance-time graph, time is taken along the horizontal axis (X-axis) and distance is taken along the vertical axis (Y-axis). This helps us see how distance changes as time passes.

Q7. A straight line sloping upward on a distance-time graph represents:
A. Object at rest
B. Uniform motion
C. Non-uniform motion
D. Object moving backwards
B. Uniform motion
When an object moves with uniform (constant) speed, the distance increases by equal amounts in equal intervals of time. On a graph, this gives a straight line that slopes upward. The slope of this line gives the speed.

Q8. A horizontal line (parallel to time axis) on a distance-time graph means:
A. Object is moving fast
B. Object is at rest
C. Object is moving slowly
D. Object is oscillating
B. Object is at rest
If the distance does not change as time increases, the line is horizontal. This means the object is not moving; it is at rest. For example, a parked car will show a horizontal line on a distance-time graph.

Q9. The slope of a distance-time graph gives:
A. Time
B. Distance
C. Speed
D. Acceleration
C. Speed
Slope is calculated as (change in distance) divided by (change in time). That is exactly the definition of speed. A steeper slope means higher speed, and a gentler slope means lower speed.

Q10. A curved line on a distance-time graph indicates:
A. Uniform motion
B. Object at rest
C. Non-uniform motion
D. No motion
C. Non-uniform motion
A curved line means the speed is changing. The object may be speeding up (increasing slope) or slowing down (decreasing slope). This type of motion is called non-uniform motion.

Q11. One complete to-and-fro motion of a pendulum bob is called:
A. One rotation
B. One oscillation
C. One vibration
D. One revolution
B. One oscillation
When the bob moves from its mean position to one extreme, back to mean, to the other extreme, and back to mean, it completes one oscillation. This is the basic unit of oscillatory motion.

Q12. The time taken to complete one oscillation is called:
A. Frequency
B. Time period
C. Amplitude
D. Speed
B. Time period
The time period is the time required for one full oscillation. It is measured in second. For example, if a swing takes 3 second to go back and forth once, its time period is 3 second.

Q13. If a pendulum completes 30 oscillations in 60 second, its time period is:
A. 0.5 second
B. 2 second
C. 30 second
D. 1800 second
B. 2 second
Time period = total time / number of oscillations = 60 second / 30 = 2 second. So each oscillation takes 2 second.

Q14. Which of the following is an example of uniform motion?
A. A train starting from a station
B. A car moving on a straight road with constant speed
C. A ball falling from a height
D. A bicycle coming to a stop
B. A car moving on a straight road with constant speed
Uniform motion means the object covers equal distances in equal intervals of time. A car moving at constant speed on a straight road does exactly that. The other options involve changing speed, so they are non-uniform.

Q15. Which of the following is an example of non-uniform motion?
A. A fan rotating at constant speed
B. A planet moving around the sun in a circular orbit at constant speed
C. A car accelerating on a highway
D. A pendulum swinging with constant time period
C. A car accelerating on a highway
Non-uniform motion means speed is changing. When a car accelerates, its speed increases with time, so it covers unequal distances in equal intervals of time. The other options involve constant speed, so they are uniform motion.

Q16. A bus covers first 50 kilometre in 1 hour and next 50 kilometre in 1.5 hour. This motion is:
A. Uniform motion
B. Non-uniform motion
C. Circular motion
D. Oscillatory motion
B. Non-uniform motion
In the first hour, speed = 50 km/h. In the next 1.5 hour, speed = 50/1.5 ≈ 33.33 km/h. Since speed is not the same throughout, the motion is non-uniform. The average speed is total distance (100 km) divided by total time (2.5 h) = 40 km/h.

Q17. A distance-time graph for an object moving with uniform speed is:
A. A curved line
B. A straight line sloping upward
C. A horizontal line
D. A vertical line
B. A straight line sloping upward
Uniform speed means distance increases equally in equal time intervals. This relationship is linear, so the graph is a straight line that goes upward as time increases.

Q18. A distance-time graph for an object at rest is:
A. A straight line sloping upward
B. A straight line sloping downward
C. A horizontal line
D. A curved line
C. A horizontal line
At rest means distance does not change even when time passes. So the distance remains constant. On a graph, this appears as a flat horizontal line parallel to the time axis.

Q19. To find the speed of an object from a distance-time graph, we calculate:
A. The area under the graph
B. The slope of the graph
C. The intercept on the distance axis
D. The length of the line
B. The slope of the graph
Slope = (change in distance) / (change in time). This ratio is exactly the speed. A steeper slope means higher speed. You can pick any two points on the straight line and divide the vertical difference by the horizontal difference.

Q20. An odometer reads 1200 kilometre at the start of a journey and 1500 kilometre at the end. The distance travelled is:
A. 300 kilometre
B. 2700 kilometre
C. 1500 kilometre
D. 1200 kilometre
A. 300 kilometre
The odometer shows total distance from the beginning. To find the distance travelled during a trip, subtract the initial reading from the final reading: 1500 km − 1200 km = 300 km.

Q21. A car travels at a speed of 15 metre per second. How much time will it take to cover 300 metre?
A. 20 second
B. 0.05 second
C. 4500 second
D. 315 second
A. 20 second
Time taken = distance / speed = 300 metre / 15 metre per second = 20 second. Remember to keep the units consistent. Here both distance and speed are in metre and metre per second, so the answer is in second.

Q22. A train covers 360 kilometre in 4 hour. What is its speed?
A. 90 km/h
B. 1440 km/h
C. 364 km/h
D. 90 m/s
A. 90 km/h
Speed = distance / time = 360 km / 4 h = 90 km/h. This is the average speed if the train did not move at exactly the same speed throughout the journey.

Q23. A bar graph is most suitable for showing:
A. Change of speed over time
B. Comparison of distances travelled by different vehicles
C. Path of a pendulum
D. Shape of a distance-time graph
B. Comparison of distances travelled by different vehicles
Bar graphs are ideal for comparing data across different categories, such as the distances covered by a car, bus, bicycle, and train in one hour. Each vehicle gets a separate bar, and the height shows the distance.

Q24. In a bar graph, the width of all bars should be:
A. Different
B. Equal
C. Increasing
D. Decreasing
B. Equal
To make a bar graph fair and easy to read, all bars must have the same width. Only the height (or length) of the bars changes according to the value of the data. Unequal widths can mislead the viewer.

Q25. On a distance-time graph, if the line becomes steeper, it means:
A. Speed is decreasing
B. Speed is increasing
C. Object is at rest
D. Object is moving backwards
B. Speed is increasing
A steeper slope means a larger change in distance in the same amount of time. That means the object is moving faster. If the slope keeps increasing, the object is accelerating (non-uniform motion).

Q26. On a distance-time graph, if the line becomes less steep, it means:
A. Speed is increasing
B. Speed is decreasing
C. Object is at rest
D. Object is moving backwards
B. Speed is decreasing
A less steep slope means a smaller change in distance in the same amount of time. That means the object is slowing down. The speed is decreasing.

Q27. The number of oscillations per second is called:
A. Time period
B. Frequency
C. Amplitude
D. Speed
B. Frequency
Frequency tells how many complete oscillations happen in one second. If a pendulum makes 5 oscillations in one second, its frequency is 5 hertz (Hz). Time period is the reciprocal of frequency.

Q28. If the frequency of a pendulum is 2 hertz, its time period is:
A. 2 second
B. 0.5 second
C. 4 second
D. 1 second
B. 0.5 second
Time period = 1 / frequency. If frequency = 2 Hz, then time period = 1/2 = 0.5 second. This means one oscillation takes half a second.

Q29. A distance-time graph for non-uniform motion is:
A. A straight line
B. A curved line
C. A horizontal line
D. A vertical line
B. A curved line
When speed changes continuously, distance does not increase at a constant rate. The graph becomes a curve. The shape of the curve tells us whether the object is speeding up (curve bending upward) or slowing down (curve bending downward).

Q30. A person walks 10 metre in 2 second, then 20 metre in 4 second, then 30 metre in 6 second. This motion is:
A. Uniform motion
B. Non-uniform motion
C. Oscillatory motion
D. Circular motion
A. Uniform motion
Check the speed in each part: first part speed = 10/2 = 5 m/s, second part speed = 20/4 = 5 m/s, third part speed = 30/6 = 5 m/s. Since speed is constant (5 m/s) throughout, the motion is uniform.

Q31. A girl covers 100 metre in 20 second, then 100 metre in 30 second. Her average speed for the whole journey is:
A. 5 m/s
B. 4 m/s
C. 6 m/s
D. 10 m/s
B. 4 m/s
Total distance = 100 m + 100 m = 200 m. Total time = 20 s + 30 s = 50 s. Average speed = total distance / total time = 200 m / 50 s = 4 m/s.

Q32. The slope of a distance-time graph for an object at rest is:
A. Zero
B. One
C. Infinite
D. Negative
A. Zero
For an object at rest, distance does not change as time changes. So change in distance = 0. Slope = 0 / (change in time) = 0. A horizontal line has zero slope.

Q33. A car moves with a speed of 25 metre per second. How much distance will it cover in 40 second?
A. 1.6 metre
B. 1000 metre
C. 65 metre
D. 15 metre
B. 1000 metre
Distance = speed × time = 25 m/s × 40 s = 1000 metre. This is equal to 1 kilometre. So the car covers 1 km in 40 second at this speed.

Q34. The odometer of a car reads 2000 kilometre at 10:00 AM and 2400 kilometre at 11:00 AM. The average speed of the car in this time is:
A. 400 km/h
B. 200 km/h
C. 40 km/h
D. 4000 km/h
A. 400 km/h
Distance travelled = 2400 km − 2000 km = 400 km. Time taken = 1 hour. Average speed = 400 km / 1 h = 400 km/h. This is a very high speed, possible only for racing cars or aeroplanes.

Q35. Which of the following is NOT shown by a distance-time graph?
A. Whether the object is at rest
B. Whether the motion is uniform or non-uniform
C. The colour of the moving object
D. The speed of the object
C. The colour of the moving object
A distance-time graph shows only the relationship between distance and time. It tells us about motion, rest, speed, and uniformity. It cannot show physical properties like colour, shape, or material of the object.

Q36. A bus travels at 30 kilometre per hour. How long will it take to cover 15 kilometre?
A. 0.5 hour
B. 2 hour
C. 450 hour
D. 30 hour
A. 0.5 hour
Time = distance / speed = 15 km / 30 km/h = 0.5 hour. Half an hour is 30 minute. This formula works for any units as long as they are consistent.

Q37. The motion of a simple pendulum is:
A. Uniform motion
B. Non-uniform motion
C. Oscillatory motion
D. Rectilinear motion
C. Oscillatory motion
A pendulum bob moves back and forth repeatedly about a mean position. This to-and-fro motion is called oscillatory motion. It is also periodic because it repeats after a fixed time interval.

Q38. On a distance-time graph, a straight line sloping downward would indicate:
A. Object moving forward
B. Object moving backward towards the start
C. Object at rest
D. Object moving with increasing speed
B. Object moving backward towards the start
If distance decreases as time increases, the object is returning toward the starting point. The slope is negative. In most simple cases, we consider only forward motion, so downward slopes are rare in basic distance-time graphs.

Q39. To compare the distances covered by five different animals in one hour, the best graph is:
A. Line graph
B. Bar graph
C. Pie chart
D. Distance-time graph
B. Bar graph
A bar graph is best for comparing quantities across different categories. Here each animal is a category, and the distance covered is the value. Bars of equal width with different heights make comparison easy.

Q40. A man runs 200 metre in 25 second, then stops for 10 second, then runs another 100 metre in 20 second. The motion is:
A. Uniform
B. Non-uniform
C. Circular
D. Oscillatory
B. Non-uniform
Speed in first part = 200/25 = 8 m/s. Then speed becomes zero during the stop (0 m/s). Then speed in last part = 100/20 = 5 m/s. Since speed changes, the overall motion is non-uniform.

Q41. The formula relating speed, distance and time is:
A. Speed = Distance × Time
B. Speed = Distance / Time
C. Time = Speed × Distance
D. Distance = Time / Speed
B. Speed = Distance / Time
This is the basic definition of speed. From this formula, we can also write Distance = Speed × Time and Time = Distance / Speed. All three forms are useful depending on what we need to find.

Q42. A car moves with uniform speed and covers 240 metre in 12 second. Its speed is:
A. 20 m/s
B. 2880 m/s
C. 0.05 m/s
D. 252 m/s
A. 20 m/s
Speed = distance / time = 240 metre / 12 second = 20 metre per second. Uniform speed means this value is constant throughout the motion.

Q43. The time taken to complete 10 oscillations of a pendulum is 25 second. The time period is:
A. 2.5 second
B. 0.4 second
C. 250 second
D. 35 second
A. 2.5 second
Time period = total time / number of oscillations = 25 second / 10 = 2.5 second. This means each oscillation takes two and a half seconds.

Q44. A distance-time graph helps us to:
A. Find the colour of the object
B. Find the speed of the object at any time
C. Find the weight of the object
D. Find the temperature of the object
B. Find the speed of the object at any time
From a distance-time graph, we can calculate speed as the slope. For uniform motion, the slope is constant. For non-uniform motion, the slope at any point gives the instantaneous speed.

Q45. If an object covers 50 metre in first 5 second, 100 metre in next 5 second, and 150 metre in next 5 second, the motion is:
A. Uniform
B. Non-uniform (accelerated)
C. Oscillatory
D. Circular
B. Non-uniform (accelerated)
Speed in first 5 s = 50/5 = 10 m/s. Speed in next 5 s = 100/5 = 20 m/s. Speed in last 5 s = 150/5 = 30 m/s. Speed is increasing, so motion is non-uniform and the object is accelerating.

Q46. A train travels at 108 kilometre per hour. Its speed in metre per second is:
A. 30 m/s
B. 388.8 m/s
C. 10.8 m/s
D. 108 m/s
A. 30 m/s
To convert km/h to m/s, multiply by 5/18. 108 × (5/18) = 108 × 5 ÷ 18 = 540 ÷ 18 = 30 m/s. This is a common conversion in physics problems.

Q47. The device that measures the speed of a vehicle at any instant is:
A. Odometer
B. Speedometer
C. Barometer
D. Thermometer
B. Speedometer
A speedometer shows the instantaneous speed of the vehicle in kilometre per hour or mile per hour. An odometer shows total distance travelled. Do not confuse the two.

Q48. On a distance-time graph, the steeper the line, the:
A. Slower the speed
B. Faster the speed
C. Longer the time
D. Shorter the distance
B. Faster the speed
Steepness (slope) is change in distance divided by change in time. A larger slope means more distance covered in the same time, which means higher speed.

Q49. A bicycle moves with a speed of 5 metre per second. How much time will it take to cover 1 kilometre?
A. 200 second
B. 0.005 second
C. 5000 second
D. 5 second
A. 200 second
First convert 1 kilometre into metre: 1 km = 1000 metre. Then time = distance / speed = 1000 metre / 5 metre per second = 200 second. That is 3 minute and 20 second.

Q50. A swinging child on a swing completes 15 oscillations in 30 second. The time period is:
A. 0.5 second
B. 2 second
C. 15 second
D. 450 second
B. 2 second
Time period = total time / number of oscillations = 30 second / 15 = 2 second. The child takes 2 second to swing forward and back once. This is an example of oscillatory motion in daily life.