Motion And Time
Class 7th Science NCERT Exemplar Solution
- Which of the following cannot be used for measurement of time?A. A leaking tap. B. Simple…
- Two clocks A and B are shown in Figure 13.1. Clock A has an hour and a minute hand,…
- Two students were asked to plot a distance-time graph for the motion described in Table A…
- A bus travels 54 km in 90 minutes. The speed of the bus isA. 0.6 m/s B. 10 m/s C. 5.4 m/s…
- If we denote speed by S, the distance by D and time by T, the relationship between these…
- Observe Figure 13.3. The time period of a simple pendulum is the time taken by it to…
- Fig. 13.4 shows an oscillating pendulum. Time taken by the bob to move from A to C is t 1…
- The correct symbol to represent the speed of an object isA. 5 m/s B. 5 mp C. 5 m/s-1 D. 5…
- Boojho walks to his school which is at a distance of 3 km from his home in 30 minutes. On…
- A simple pendulum is oscillating between two points A and B as shown in Figure 13.5. Is…
- Paheli and Boojho have to cover different distances to reach their school but they take…
- If Boojho covers a certain distance in one hour and Paheli covers the same distance in two…
- Complete the data of the table given below with the help of the distance-time graph given…
- The average age of children of Class VII is 12 years and 3 months. Express this age in…
- A spaceship travels 36,000 km in one hour. Express its speed in km/s.…
- Starting from A, Paheli moves along a rectangular path ABCD as shown in Figure 13.7. She…
- Plot a distance-time graph of the tip of the second hand of a clock by selecting 4 points…
Multiple Choice Questions
Question 1.Which of the following cannot be used for measurement of time?
A. A leaking tap.
B. Simple pendulum.
C. Shadow of an object during the day.
D. Blinking of eyes.
Answer:
Blinking of eyes cannot be used for measurement of time because it depends upon person to person and also it is not a constant event.
Question 2.
Two clocks A and B are shown in Figure 13.1. Clock A has an hour and a minute hand, whereas clock B has an hour hand, minute hand as well as a second hand. Which of the following statement is correct for these clocks?
A. A time interval of 30 seconds can be measured by clock A.
B. A time interval of 30 seconds cannot be measured by clock B.
C. Time interval of 5 minutes can be measured by both A and B.
D. Time interval of 4 minutes 10 seconds can be measured by clock A.
Answer:
As both, the clock is having a minute hand so we can measure 5 minutes.
Since Clock A doesn’t have a second hand so it can’t measure time in second and thus other options are not correct.
Question 3.
Two students were asked to plot a distance-time graph for the motion described in Table A and Table B. Table A Table B.
The graph given in Figure 13.2 is true for
A. both A and B.
B. An only.
C. B only.
D. neither A nor B.
Answer:
For table A: we mark all the point of time and check what is the distance then we compare from the table. Look here:-
Graph: Table1
For the second table, we look for time in points given in the table and then draw a line cutting x-axis from that point and parallel to y-axis. This line cuts the dotted line at some point, this point is required a point to be plotted on the graph corresponding to point given in the table. Also, this point can be plotted with y-axis coordinate given in the table.
After plotting all points graph looks like:
Graph: Table2
Question 4.
A bus travels 54 km in 90 minutes. The speed of the bus is
A. 0.6 m/s
B. 10 m/s
C. 5.4 m/s
D. 3.6 m/s
Answer:
Distance = 54 km
= 54 × 1000 m
Time = 90 min
= 90 × 60 sec
Speed = 
= 
= 
= 10 m/s
Question 5.
If we denote speed by S, the distance by D and time by T, the relationship between these quantities is
A. S = D × T
B. 
C. 
D. 
Answer:
S = 
Question 6.
Observe Figure 13.3.
The time period of a simple pendulum is the time taken by it to travel from
A. A to B and back to A.
B. O to A, A to B and B to A.
C. B to A, A to B and B to O.
D. A to B.
Answer:
Time period is defined as the time taken by the pendulum to complete one oscillation. Here A to B and again back to A complete one oscillation.
Question 7.
Fig. 13.4 shows an oscillating pendulum.
Time taken by the bob to move from A to C is t 1 and from C to O is t 2 . The time period of this simple pendulum is
A. (t 1 + t 2)
B. 2 (t 1 + t 2)
C. 3 (t 1 + t 2)
D. 4 (t 1 + t 2)
Answer:
Time period is time taken by the pendulum to complete one oscillation.
one oscillation =
time taken from A to O and O to B is the same.
One oscillation = (A to C) + (C to O) + (O to B) + (O to C) + (C to A)
= (t1 + t2) + (t1 + t2) + (t1 + t2)+ (t1 + t2) = 4(t1 + t2)
Question 8.
The correct symbol to represent the speed of an object is
A. 5 m/s
B. 5 mp
C. 5 m/s-1
D. 5 s/m
Answer:
speed = distance / time
Distance is represented in metre (m) and time in second (s), so speed become m/s. Hence option (a) is correct.
Question 9.
Boojho walks to his school which is at a distance of 3 km from his home in 30 minutes. On reaching he finds that the school is closed and comes back by bicycle with his friend and reaches home in 20 minutes. His average speed in km/h is
A. 8.3
B. 7.2
C. 5
D. 3.6
Answer:
Boojho Home to school to again home
From home to school distance is 3 km time taken is 30 min = (30/60) hour = 0.5 hour
And in returning from school to home time taken = 20 min = (20/60) hour = 0.33 hour
average speed = Total distance / time taken
= (3 + 3)/(0.5 + 0.33) km/h
= 7.2 km/h
Very Short Answer Questions
Question 1.A simple pendulum is oscillating between two points A and B as shown in Figure 13.5. Is the motion of the bob uniform or non-uniform?
Answer:
A simple pendulum is oscillating between two points A and B is uniform motion. As Bob (small metallic ball) moves in a constant speed throughout the motion so it is a uniform motion.
Question 2.
Paheli and Boojho have to cover different distances to reach their school but they take the same time to reach the school. What can you say about their speed?
Answer:
As Paheli and Boojho both reach school at the same time but their distance to reach school is different so we can say that their speed is different. The one who covers more distance has more speed than those who cover less.
Question 3.
If Boojho covers a certain distance in one hour and Paheli covers the same distance in two hours, who travels in a higher speed?
Answer:
Let the distance cover by Boojho be x km
So, the speed of Boojho will be = 
km/h
Paheli also cover x km, so her speed will be = 
km/h
So, we can see that Boojho travels with higher speed higher than Paheli.
Short Answer Questions
Question 1.Complete the data of the table given below with the help of the distance-time graph given in Figure 13.6.
Answer:
We have to fill 3 blanks
First is the distance at time 4 second,
Second is time at distance 12 m and the last one is the distance at time 8 second.
The distance at time 4 second: Draw a line from 4 sec (pt A) at x-axis parallel to y-axis mark the point where it intersects the graph (pt B) draw a line parallel to x-axis from pt B and it interest y-axis at pt C. OC represent distance in 4 sec, distance is 8 m.
Similarly, we do for finding time at distance 12 m OD is a time in 12 m, so OC = 6 seconds.
Also for the third part distance in 8 sec, OI represent distance in 8 sec. OI is 16 m. So your table will look like:-
Question 2.
The average age of children of Class VII is 12 years and 3 months. Express this age in seconds.
Answer:
Average age is 12 years and 3 months
1 year is having 365 days (assuming only in leap year)
1 day is having 24 hour
1 hour is 60min
1 min is 60 second
Now doing all these in reverse order,
1 hour = 60×60 second
1 day = 24×60×60 second
1 year = 365×24×60×60 second
So, 12 year = 12×365×24×60×60 second = 378432000 sec
1 months = 30 days
3 months will be 3×30 days
= 3×30×24×60×60 second = 7776000 sec
12 years and 3 months = 378432000 + 7776000
= 386208000 second
Question 3.
A spaceship travels 36,000 km in one hour. Express its speed in km/s.
Answer:
Spaceship speed is 36000 km/h
So, in 1 hour spaceship travel 36000 km
Or we can say that in 60×60 second spaceship travel 36000 km
Therefore in 1 sec spaceship travel
= 
= 10 km/s
Question 4.
Starting from A, Paheli moves along a rectangular path ABCD as shown in Figure 13.7. She takes 2 minutes to travel each side. Plot a distance-time graph and explain whether the motion is uniform or non-uniform.
Answer:
Here is a graph between distance and time:
x-axis represents time in minutes and y-axis represent distance in the meter.
Paheli starts from position A and go to point B in 2 min so distance cover is 8 m.We mark point (A) 2 min on axis and point (C) on y-axis then we draw a line passing to pt(A) parallel to the y-axis and one more line passing to pt(C) parallel to the x-axis. Where the two-line interest will be marked as pt(B). Join the pt(B) and pt(O). This line represents a graph of one side
Similarly, we do for all the side.
Speed of Paheli will moving alongside AB is (distance/time) = (
) m/min = 4 m/min
Here the unit is meter/minute.
Speed along side BC is = 
= 3 m/min
Speed alongside CD is = 
= 4 m/min
Speed along side DA is = = 3 m/min
So, we find that Paheli speed is not the same in the whole journey.
Motion is nonuniform.
Nonuniform can also be found from the distance-time graph, here line is not straight so it is non-uniform motion.
Question 5.
Plot a distance-time graph of the tip of the second hand of a clock by selecting 4 points on the x-axis and y-axis respectively. The circumference of the circle traced by the second hand is 64 cm.
Answer:
Before plotting the graph, we take 4 points on 3, 6, 9 and 12 hours of a clock.
Point A: second hand cover 16 cm and take 15 seconds to reach at this point.
Point B: here second hand take 30 seconds and cover 32 cm.
Similarly, for point C: 45 second and 48 cm and point D: 60 second and 64 cm
Below are table, clock and graph -
Long Answer Questions
Question 1.Given below as Figure 13.8 is the distance-time graph of the motion an object.
(i) What will be the position of the object at the 20s?
(ii) What will be the distance travelled by the object in 12s?
(iii) What is the average speed of the object?
Answer:
i) So to find position of object at time 20 second, we draw line parallel to y-axis from time 20 and take the point where it intersect then again draw line parallel to x-axis, we get the distance where it intersect the y- axis.
Here position is 8 meter from starting.
ii) Distance travel in 12 second by the object is 6 meter.
iii) Since average speed of object is define as total distance covered by object by total time take. So here total distance is 8 m and total time is 20 second,
Average speed = 
= 0.4 m/s
Question 2.
Distance between Bholu’s and Golu’s house is 9 km. Bholu has to attend Golu’s birthday party at 7 o’clock. He started from his home at 6 o’clock on his bicycle and covered a distance of 6 km in 40 minutes. At that point he met Chintu and he spoke to him for 5 minutes and reached Golu’s birthday party at 7 o’clock. With what speed did he cover the second part of the journey? Calculate his average speed for the entire journey.
Answer:
Party start at 7’o clock
Total time spend by Bholu = 40 min (traveling) + 5 min (chatting) = 45 min
So Bholu has to reach Golu’s house within 15 min (1hour – 45 min)
The speed with which Bholu has to cover the second part of the journey is
= 
= 
Average The average of Bholu’s in the entire journey is total distance divided by total the time taken
= 
= 
Question 3.
Boojho goes to the football ground to play football. The distance-time graph of his journey from his home to the ground is given as Figure 13.9.
(a) What does the graph between point B and C indicate about the motion of Boojho?
(b) Is the motion between 0 to 4 minutes uniform or nonuniform?
(c) What is his speed between 8 and 12 minutes of his journey?
Answer:
a) Graph between point B and C is parallel to the x-axis as time is changing but distance remains the same means there is no motion.
Boojho is at rest position.
b) Motion between time 0 min to 4 min is nonuniform because the graph is not straight. It like two-line joining at point A.
c) At time 12-minute distance is 225 m and at time 8-minute distance is 150 m so speed is distance divided by the time taken
Speed = 
= 18.75 m/min