Motion and Time NCERT Solutions – Class 7 Science

Q.1 Classify the following as motion along a straight line, circular or oscillatory motion:
(i) Motion of your hands while running.
(ii) Motion of a horse pulling a cart on a straight road.
(iii) Motion of a child in a merry-go-round.
(iv) Motion of a child on a see-saw.
(v) Motion of the hammer of an electric bell.
(vi) Motion of a train on a straight bridge.

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Q.2 Which of the following are not correct?
(i) The basic unit of time is second.
(ii) Every object moves with a constant speed.
(iii) Distances between two cities are measured in kilometres.
(iv) The time period of a given pendulum is constant.
(v) The speed of a train is expressed in m/h.
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Following statements are not correct:

(ii) Every object moves with a constant speed.

(v) The speed of a train is expressed in m/h.

Q.3 A simple pendulum takes 32 s to complete 20 oscillations. What is the time period of the pendulum?

$\begin{array}{l}\text{<strong>Ans-</strong>}\end{array}$ $\begin{array}{l}\text{Time period of a simple pendulum is the time taken to complete one oscillation.}\\ \text{The time period is given by,}\\ \text{Time period =}\frac{\text{Time taken}}{\text{Number of oscillations}}=\frac{32\mathrm{s}}{20}=1.6\mathrm{s}\end{array}$

Q.4 The distance between two stations is 240 km. A train takes 4 hours to cover this distance. Calculate the speed of the train.

\begin{array}{l}\text{Distance travelled = 240 km}\\ \text{Time taken = 4 hours}\\ \text{Let the speed of the train be x km/h}\\ \text{Speed =}\frac{\mathrm{distance}}{\text{time taken}}\\ \therefore \text{speed of train =}\frac{240\text{km}}{4\text{h}}=\text{60 km/h}\end{array}

Q.5 The odometer of a car reads 57321.0 km when the clock shows the time 08:30 AM. What is the distance moved by the car, if at 08:50 AM, the odometer reading has changed to 57336.0 km? Calculate the speed of the car in km/min during this time. Express the speed in km/h also.

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\begin{array}{l}\text{Difference in the readings of odometer =}\left(\text{final – initial}\right)\text{readings = distance travelled}\\ \text{= 57336}\text{.0 km = 15 km}\\ \text{Difference in readings of clock = time taken = 08}\text{.50 AM – 08}\text{.30 AM = 20 min}\\ \therefore \text{the speed of car in km/min =}\frac{\text{distance}}{\text{time taken}}\text{=}\frac{\text{15 km}}{\text{20 min}}\text{= 0}\text{.75 km / min}\\ \text{The speed of the car in km/h =}\frac{\text{15 km}}{\text{1/3h}}\text{= 45 km/h}\end{array}

Q.6 Salma takes 15 minutes from her house to reach her school on a bicycle. If the bicycle has a speed of 2 m/s, calculate the distance between her house and the school.
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Time taken by Salma = 15 min = 15 × 60 s = 900 s

Speed of bicycle = 2 m/s

Let the distance between school and house be x m

Distance travelled = time taken × speed

∴ Distance, x = 900 s × 2 m/s = 1800 m = 1.8 km

Sr9 (1674867)

Q.7 Show the shape of the distance-time graph for the motion in the following cases:
(i) A car moving with a constant speed.
(ii) A car parked on a side road.

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(i) The shape of the distance-time graph for the motion of car with constant speed will be as follows:

(ii) The shape of the distance-time graph for the motion of car parked on a side road will be as follows:

$\begin{array}{l}\mathrm{<strong>Q.8\; </strong>Which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{relation}\mathrm{is}\mathrm{correct}?\end{array}$ $\begin{array}{l}\left(\text{(i)}\right)\mathrm{speed}=\mathrm{Distance}\times \mathrm{time}\end{array}$ $\begin{array}{l}\left(\mathrm{(ii)}\right)\mathrm{speed}=\frac{\mathrm{Distance}}{\mathrm{time}}\end{array}$ $\begin{array}{l}\left(\mathrm{(iii)}\right)\mathrm{speed}=\frac{\mathrm{time}}{\mathrm{Distance}}\\ \left(\mathrm{iv}\right)\mathrm{speed}=\frac{1}{\mathrm{Distance}\times \mathrm{time}}\end{array}$

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Correct option is (ii).

Q.9 Explanation: Speed is the distance traveled per unit of time.

The basic unit of speed is:
(i) km/min
(ii) m/min
(iii) km/h
(iv) m/s

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Correct option is (iv).
Explanation: Speed is the distance traveled per unit of time. The standard unit of distance is metre (m) and standard unit of time is second (s).

Q.10 A car moves with a speed of 40 km/h for 15 minutes and then with a speed of 60 km/h for the next 15 minutes. The total distance covered by the car is:
(i) 100 km
(ii) 25 km
(iii) 15 km
(iv) 10 km

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Average speed of car = (40 km/h + 60 km/h)/2 = 50 km/h
Total distance travelled = speed × total time taken = 50 km/h × (½) h = 25 km/h

Q.11 Suppose the two photographs, shown in Fig. 13.1 and Fig. 13.2, had been taken at an interval of 10 seconds. If a distance of 100 metres is shown by 1 cm in these photographs, calculate the speed of the blue car.

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The steps to be followed are as follows:
1.
Find the distance travelled by blue car in photographs by measuring distance taking mid-points of car in following manner:

(Here, distance may vary in centimeters due to difference in the size of photographs taken. So, follow only the method for finding right answer, which may vary.)

2. According to the question, 1 cm is equivalent to 100 m.
Therefore, distance travelled = 0.5 cm x 100m = 50m

3. Speed = distance travelled / time taken
Therefore, speed = 50m / 10s = 5 m/s

Q.14 Fig.13.15 shows the distance-time graph for the motion of two vehicles A and B. Which one of them is moving faster?

Ans-

Among the vehicles, vehicle A is moving faster than B. It is so because A has travelled greater distance in shorter period of time in comparison to that of vehicle B. Vehicle B has moved lesser distance in greater period of time as depicted in graph.

Q.13 Which of the following distance-time graphs shows a truck moving with speed which is not constant?

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Among the given distance-time graphs, following graph shows the movement of a truck with speed which is not constant:

It is so because in this graph vehicle is not covering equal distances in equal intervals of time.