NCERT Solutions Class 7 Science Chapter 9 Motion and Time

Motion and Time, the ninth chapter of Class 7 Science, covers key topics like Slow or Fast, Speed, Measurement of Time, Measuring Speed, and the Distance-Time Graph. These concepts help students understand object movement and time measurement.

For detailed explanations and exercises solutions, students can refer to NCERT Solutions for Class 7 Science Chapter Motion and Time. NCERT Solutions for Class 7 Science offer helpful insights for better preparation.

1.0Download Class 7 Science Chapter 9 NCERT Solutions PDF Online

This article offers Class 7 Motion and Time solutions aligned with NCERT guidelines. Practising these solutions helps students develop a strong foundation in Physics and improves their problem-solving abilities, ultimately aiding in achieving higher scores in exams. For more in-depth learning, students can download the NCERT Class 7 Science Chapter 9 PDF Solutions, carefully crafted by ALLEN’s experts, using the link provided below.

2.0Class 7 Science Chapter 9 Motion and Time Overview

Before diving into the Solutions for NCERT Class 7 Science Chapter Motion and Time, let’s first take a quick overview of the key topics and subtopics covered in this chapter of the NCERT Class 7 Science book.

Class 7 Science Chapter 7 - List of Topics

1. Slow or Fast: This topic explains how the speed of objects can vary and how to compare their motion.
2. Speed: It covers the concept of speed as the distance travelled per unit of time.
3. Measurement of Time: This section discusses different methods and tools used to measure time accurately.
4. Measuring Speed: It focuses on how speed is calculated and measured in different scenarios.
5. Distance-Time Graph: This explains how to represent an object's motion graphically by plotting distance against time.

3.0NCERT Questions with Solutions for Class 7 Science Chapter 9 - Detailed Solutions

• Classify the following as motion along a straight line, circular or oscillatory motion: (i) Mot ion 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. Solution (i) Oscillatory motion While running, the hands move to and fro and repeat their motion after a given interval of time. Hence, it is an oscillatory motion. (ii) Straight line (Rectilinear motion) The horse is pulling a cart on a straight road. Therefore, it has a motion along a straight line. (iii)Circular motion Merry-go-round has a circular motion. Therefore, a child sitting inside it will also have a circular motion. (iv) Oscillatory motion The child on a see-saw goes up and down continuously. It oscillates updown. Therefore, it is an oscillatory motion. (v) Oscillatory motion The hammer hits the electric bell and vibrates rapidly. Therefore, it is an oscillatory motion. (vi) Straight line(Rectilinear motion) The train is moving on a straight bridge. Therefore, it has a motion along a straight line.
• Which of the following are correct or incorrect? (i) The basic unit of time is second. (ii) Every object moves with a constant speed. (iii)Distance between two cities are measured in kilometres. (iv) The time period of a given pendulum is not constant. (v) The speed of a train is expressed in $\mathrm{m}/\mathrm{h}$. Solution (i) Correct Second is the SI unit of time. (ii) incorrect An object can move with constant or variable speed. (iii)Correct The distance between two cities can be very large. Since kilometre is a bigger unit of distance, the distance between two cities is measured in kilometres. (iv) incorrect Time period of a pendulum depends on the length of the thread. Hence, it is constant for a particular pendulum. (v) incorrect The speed of a train is measured either in $\mathrm{km}/\mathrm{h}$ or in $\mathrm{m}/\mathrm{s}$.
• A simple pendulum takes 32 s to complete 20 oscillations. What is the time period of the pendulum? Solution Number of oscillations $=20$ Total time taken to complete 20 oscillations $=32\mathrm{s}$ Time period $=\frac{\text{ Total time taken }}{\text{ Number of oscillations }}$ $=\frac{32}{20}=1.6\mathrm{S}$
• The distance between two stations is 240 km. A train takes 4 hours to cover this distance. Calculate the speed of the train. Solution Distance between the two stations $=240\mathrm{km}$ Time taken $=4\mathrm{h}$ Speed $=\frac{\text{ Distance }}{\text{ Time taken }}=\frac{240}{4}=60\mathbf{km}/\mathbf{h}$
• 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 $\mathrm{km}/\min$ during this time. Express the speed in $\mathrm{km}/\mathrm{h}$ also. Solution Initial reading of the odometer of the car $=57321.0\mathrm{km}$ Final reading of the odometer of the car=57336.0 km Distance covered by the car = Final reading of the odometer of the car - Initial reading of the odometer of the car $=$ 57336.0-57321.0 = 15 km The given car starts at 8:30 a.m. and stops at 8:50 a.m. Therefore, time taken by the car to cover the distance is (8:50-8:30) min $=20\min$ Distance covered by the car $=15\mathrm{km}$ Time taken by the car $=20\min$ Speed $=\frac{\text{ Distance covered }}{\text{ Time taken }}=\frac{15}{20}$ $=0.75\mathrm{km}/\min$ Again, $60\min =1\mathrm{h}$ $20\min =\frac{1}{60}\times 20=\frac{1}{3}\mathrm{h}$ The time taken by the car $=\frac{1}{3}\mathrm{h}$ Speed $=\frac{\text{ Distance covered }}{\text{ Time taken }}$ speed $=\frac{15}{\frac{1}{3}}=\frac{15\times 3}{1}=45\mathrm{km}/\mathrm{h}$
• Salma takes 15 minutes from her house to reach her school on a bicycle. If the bicycle has a speed of $2\mathrm{m}/\mathrm{s}$, calculate the distance between her house and the school. Solution Time taken by Salma to reach her school from her home $=15\min =15\times 60=900\mathrm{s}$ Speed of her bicycle $=2\mathrm{m}/\mathrm{s}$ Speed $=\frac{\text{ Distance covered }}{\text{ Timetaken }}$ Distance covered $=$ Speed $\times$ Time taken $=2\times 900=1800\mathrm{m}$ $1000\mathrm{m}=1\mathrm{km}$ $\therefore 1800\mathrm{m}=\frac{1}{1000}\times 1800=1.8\mathrm{km}$
• 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. Solution (i) A car moving with a constant speed covers equal distance in equal intervals of time. Such motion of car is represented in the given distance-time graph.
  
  (ii) The distance-time graph of a car parked on a road side is such that with the increase in time, there is no change in distance, as shown in the given figure.
  
• Which of the following relations is correct? (i) Speed = Distance x Time (ii) Speed $=\frac{\text{ Distance }}{\text{ Time }}$ (iii) Speed $=\frac{\text{ Time }}{\text{ Distance }}$ (iv) Speed $=\frac{1}{\text{ Distance }\times \text{ Time }}$ Solution (ii) Speed of an object is given by the relation Speed $=\frac{\text{ Distance }}{\text{ Time }}$
• The basic unit of speed is: (1) $\mathrm{km}/\min$ (2) $\mathrm{m}/\min$ (3) $\mathrm{km}/\mathrm{h}$ (4) $\mathrm{m}/\mathrm{s}$ Solution Option (4) The basic unit of distance is metre ( m ). The basic unit of time is second (s). Speed $=\frac{\text{ Distance }}{\text{ Time }}$ Therefore, the basic unit of speed is $\mathrm{m}/\mathrm{s}$.
• A car moves with a speed of $40\mathrm{km}/\mathrm{h}$ for 15 minutes and then with a speed of 60 $\mathrm{km}/\mathrm{h}$ for the next 15 minutes. The total distance covered by the car is : (1) 100 km (2) 25 km (3) 15 km (4) 10 km Solution Option (2) Case I Speed of the car $=40\mathrm{km}/\mathrm{h}$ Time taken $=15\min =\frac{15}{60}=0.25\mathrm{h}$ Speed $=\frac{\text{ Distance covered }}{\text{ Timetaken }}$ Distance covered, $d_1=$ Speed $\times$ Time taken $=40\times 0.25=10\mathrm{km}$ Case II Speed of the car $=60\mathrm{km}/\mathrm{h}$ Time taken $=15\min =\frac{15}{60}=0.25\mathrm{h}$ Speed $=\frac{\text{ Distance covered }}{\text{ Timetaken }}$ Distance covered, ${\mathrm{d}}_2=$ Speed $\times$ Time taken $=60\times 0.25=15\mathrm{km}$ Total distance covered by the car, $\mathrm{d}={\mathrm{d}}_1+{\mathrm{d}}_2=10+15=25\mathbf{km}$ Therefore, the total distance covered by the car is 25 km .
• Suppose the two photographs, shown in figure 1 and Figure 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.
  
  Vehicles moving in the same direction of on a road
  
  Position of vehicles shown in figure after some time Solution The distance covered by the blue car (as evident from the photograph) from one white strip to another, which is measured by scale is 1.4 cm . It is given that 1 cm is equivalent to 100 m . Therefore, 1.4 cm is equivalent to 140 m . Distance travelled by the car $=140\mathrm{m}$ Time interval between the two photographs $=10\mathrm{s}$ $\begin{array}{rl}\text{ Speed }& =\frac{\text{ Distance covered }}{\text{ Timetaken }}\\ & =\frac{140}{10}=14m/\mathbf{s}\end{array}$
• Figure shows the distance-time graph for the motion of two vehicles A and B. Which one of them is moving faster?
  
  Distance-time graph for the motion of two cars Solution Vehicle A is moving faster than vehicle B. Speed is given by the relation Speed $=\frac{\text{ Distance covered }}{\text{ Timetaken }}$
  
  This relation shows that speed of a vehicle is greater if it covers maximum distance in a given interval of time. To compare the distance, draw a line perpendicular to the time-axis, as shown in the following distance-time graph.
  
  From the graph, it is evident that for a given time $t$, the distance covered by vehicle A is more than vehicle B. Hence, vehicle $A$ is moving faster than vehicle $B$.
• Which of the following distance-time graphs shows a truck moving with speed which is not constant?
  
  (i)
  
  (iii)
  
  (ii)
  
  (iii) Solution Graph (iii) In a distance-time graph, the constant speed of a truck will be represented by a straight line. In a distance-time graph, a straight line parallel to the time axis indicates that the truck is not moving. A curved line on the distance-time graph indicates that the truck is moving with a speed which is not constant.

4.0Benefits of NCERT Solutions Class 7 Science Chapter 7 

Studying Class 7 Science Chapter 9(Motion and Time) offers several benefits:

1. Understanding Basic Physics Concepts: It introduces fundamental concepts like speed, time, and motion, forming the basis for future studies in physics.
2. Improved Problem-Solving Skills: Students learn how to calculate speed, measure time, and interpret distance-time graphs, enhancing analytical and problem-solving abilities.
3. Practical Application: The concepts learned are applicable in everyday life, helping students better understand how things move and time is measured.
4. Graphical Representation: Learning to interpret distance-time graphs helps students visualize data and understand the relationship between distance and time.
5. Foundation for Higher Studies: This chapter builds a strong foundation for advanced topics in physics, making it easier to grasp more complex concepts in the future.