Class 7 Science Motion and Time Chapter Notes

Class 7 Science Motion and Time Chapter Notes. There are many more useful educational material which the students can download in pdf format and use them for studies. Study material like concept maps, important and sure shot question banks, quick to learn flash cards, flow charts, mind maps, teacher notes, important formulas, past examinations question bank, important concepts taught by teachers. Students can download these useful educational material free and use them to get better marks in examinations.  Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Motion and Time

3.1 Introduction

In our daily life, we see various objects in motion. The walking man, the running cars, the water flowing in lakes, flying aeroplanes etc. are all examples of objects in motion.

3.2 Rest and Motion

Rest: A body is said to be at rest if it does not change its position with respect to time with respect to another body. Example: a book lying on a table.Motion: A body is said to be in motion if it changes its position with respect to time with respect to another body. Example: a train moving on rails, a man walking on a road.

3.3 Physical quantities

Physical quantity: The quantity which can be measured is called a physical quantity. In Physics, we study physical quantities.

Types of physical quantity:

1. Fundamental physical quantity
2. Derived physical quantity

Fundamental physical quantity: The physical quantities which are independent of other physical quantities are called fundamental physical quantities. Example: Length, Mass, Time etc.Derived physical quantity: The physical quantities which depend on fundamental physical quantities are called derived physical quantities. Example: Speed, Energy etc.

All physical quantities can be categorised into scalars and vectors.

Scalars: The quantities which have magnitude only are called scalars. Example: Distance, Speed, Energy etc.Vectors: The quantities which have both magnitude and direction are called vectors. Examples: Displacement, Velocity, Force etc. A vector is represented by putting an arrow over the symbol of the physical quantity; for example, velocity is represented as \(\vec{v}\), force is represented as \(\vec{F}\), etc.

3.4 Speed

We know that some vehicles move faster than others. The distance moved by objects in a given interval of time can help us to decide which one is faster or slower.

Speed can be defined as the distance covered by an object in unit time. Thus

\[ \text{Speed}=\frac{\text{Total distance covered}}{\text{Total time taken}} \]

The S.I. unit of speed is m/s and it is a scalar quantity.

3.5 Velocity

The velocity of an object is the ratio of total displacement to the total time taken. Thus

\[ \text{Velocity}=\frac{\text{Total displacement}}{\text{Total time taken}} \]

The S.I. unit of velocity is the same as that of speed, m/s, but velocity is a vector quantity, because it also determines the direction of motion.

3.6 Acceleration

The acceleration of a moving object is defined as the rate of change in its velocity with time. Thus

\[ \text{Acceleration}=\frac{\text{Change in velocity}}{\text{Total time taken}}=\frac{\vec{v}_2-\vec{v}_1}{t} \]

The S.I. unit of acceleration is m/s² and it is a vector quantity.

3.7 Types of motion

(a) Uniform Motion: If an object is moving along a straight line with constant speed/velocity, it is said to be in uniform motion.

For uniform motion, velocity = constant, i.e. \(\vec{v}_2=\vec{v}_1=\vec{v}\) (let us say). Then

\[ \vec{a}=\frac{\vec{v}_2-\vec{v}_1}{t}=\frac{\vec{v}-\vec{v}}{t}=0 \]

So, for uniform motion the acceleration is 0.

Equation of motion for uniform motion:

Distance/Displacement = speed × time = velocity × time

Note: For uniform motion along a straight line, speed and velocity are the same. Similarly, distance and displacement are also the same for the above mentioned case.(b) Non-uniform motion: If the velocity of an object keeps on changing with time, the motion is said to be non-uniform. Since the velocity of the object is not constant, it possesses acceleration. So, depending on the acceleration, motion can be of two types:

1. Uniformly accelerated motion: where the acceleration of the body remains constant.
2. Non-uniformly accelerated motion: where the acceleration of the body keeps on changing with time.

Equations of motion for a uniformly accelerated body:

(a) Horizontal motion

\[ \text{(i) } v=u+at \qquad \text{(ii) } s=ut+\tfrac{1}{2}at^2 \]

\[ \text{(iii) } v^2=u^2+2as \qquad \text{(iv) } s_n=u+\frac{a}{2}(2n-1) \]

where v = final velocity, u = initial velocity, a = constant acceleration, s = displacement, t = time, and s_n = displacement in the n^th second.

(b) Vertical motion

\[ \text{(i) } v=u+gt \qquad \text{(ii) } s=ut+\tfrac{1}{2}gt^2 \qquad \text{(iii) } v^2=u^2+2gh \]

\[ \text{(iv) } s_n=u+\frac{g}{2}(2n-1) \]

where g = acceleration due to gravity = 9.8 m/s² or 10 m/s².

3.8 Chart of Rest and motion

3.9 Distance time graph

The distance-time graph is a line graph. Let us learn to make such a graph. Take a sheet of graph paper. Draw two lines perpendicular to each other on it, as shown in the figure given below:

Fig. 13.1: x-axis and y-axis on a graph paper

Mark the horizontal line as XOX′. It is known as the x-axis. Similarly mark the vertical line YOY′. It is called the y-axis. The point of intersection of XOX′ and YOY′ is known as the origin O. You can make the graph by following the steps given below:

Draw two perpendicular lines to represent the two axes and mark them as OX and OY as in the given figure. Decide the quantity to be shown along the x-axis and that to be shown along the y-axis. In this case we show the time along the x-axis and the distance along the y-axis. Choose a scale to represent the distance and another to represent the time on the graph. For the motion of the car the scales could be

Time: 1 min = 1 cm. Distance: 1 km = 1 cm. Mark values for the time and the distance on the respective axes according to the scale you have chosen. Now you have to mark the points on the graph paper to represent each set of values for distance and time. The corresponding figure shows the set of points on the graph corresponding to the positions of the car at various times.

Fig. 13.2: Making a graph

Join all the points on the graph. This is the distance-time graph for the motion of the car. If the distance-time graph is a straight line, it indicates that the object is moving with a constant speed. However, if the speed of the object keeps changing, the graph can be of any shape.

EXERCISES

Objective Questions

The distance between two stations is 240 km. A train takes 6 hours to cover this distance. The speed of the train is

1. 60 km/hr
2. 40 km/hr
3. 30 km/hr
4. 24 km/hr

Answer: (b) 40 km/hr. Speed = distance ÷ time = 240 ÷ 6 = 40 km/hr.

The S.I. unit of acceleration is

1. m/s
2. ms
3. m/s²
4. m² s

Answer: (c) m/s².

A car moves with a speed of 30 km/hr for 20 minutes. The total distance covered by the car is

1. 25 km
2. 10 km
3. 30 km
4. 50 km

Answer: (b) 10 km. Time = 20 min = 1/3 hr, so distance = speed × time = 30 × 1/3 = 10 km. (The printed answer key gave (a) 25 km, which is not correct.)

A car travels between two points A and B such that it takes 20 minutes from A to B and 10 minutes from B to A. If displacement between A to B is 20 m then average velocity in the whole journey is

1. 10 m/s
2. 20 m/s
3. 15 m/s
4. zero

Answer: (d) zero. The car returns to its starting point, so the total displacement is zero and therefore the average velocity is zero.

A car changes its velocity from 10 m/s to 15 m/s in 5 sec. The acceleration of the car is

1. 1 m/s²
2. 2 m/s²
3. 3 m/s²
4. 5 m/s²

Answer: (a) 1 m/s². a = (15 − 10) ÷ 5 = 1 m/s².

Which of the following is a vector quantity ?

1. Distance
2. Speed
3. Velocity
4. Mass

Answer: (c) Velocity, because it has both magnitude and direction.

A car starts from rest and acquires velocity 10 m/s in 5 sec. The acceleration of the car is

1. 10 m/s²
2. 2 m/s²
3. 20 m/s²
4. 1 m/s²

Answer: (b) 2 m/s². a = (10 − 0) ÷ 5 = 2 m/s².

A car is moving with velocity 20 m/s. What will be its final velocity after 5 sec if it has acceleration 2 m/s² ?

1. 15 m/s
2. 20 m/s
3. 25 m/s
4. 30 m/s

Answer: (d) 30 m/s. v = u + at = 20 + 2 × 5 = 30 m/s.

A simple pendulum takes 20 seconds to complete 5 oscillations. The time period of the pendulum is

1. 2 sec
2. 3 sec
3. 4 sec
4. 5 sec

Answer: (c) 4 sec. Time period = total time ÷ number of oscillations = 20 ÷ 5 = 4 s. (The printed question said "5 seconds to complete 20 oscillations", which would give 0.25 s, not matching any option; the answer key (c) 4 sec fits 20 s for 5 oscillations.)

Subjective Questions

Define rest and motion. Give two examples of each.

Answer: A body is at rest if it does not change its position with time with respect to another body. Examples: a book lying on a table; a parked car. A body is in motion if it changes its position with time with respect to another body. Examples: a train moving on rails; a man walking on a road.

Define scalars and vectors. Give two examples of each.

Answer: Scalars are quantities which have magnitude only, e.g. distance and speed. Vectors are quantities which have both magnitude and direction, e.g. displacement and velocity.

A car starts from rest and acquires a velocity 20 m/s in 5 sec. Find the distance travelled by it during this time interval.

Answer: Given u = 0, v = 20 m/s, t = 5 s. Acceleration a = (v − u)/t = 20/5 = 4 m/s². Distance s = ut + ½at² = 0 + ½ × 4 × 5² = 50 m.

Define acceleration. Write its unit.

Answer: Acceleration is the rate of change of velocity with time, i.e. acceleration = change in velocity ÷ time taken. Its S.I. unit is m/s².

Priya takes 20 minutes from her house to reach her school on a bicycle. If bicycle has speed 2 m/s, calculate distance between her house and school.

Answer: Time = 20 min = 20 × 60 = 1200 s. Distance = speed × time = 2 × 1200 = 2400 m = 2.4 km.

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.

Answer: Distance moved = 57336.0 − 57321.0 = 15.0 km. Time taken = 08:50 − 08:30 = 20 min. Speed = 15 ÷ 20 = 0.75 km/min. In km/h: 0.75 × 60 = 45 km/h.

Draw a distance time graph from the following data

Distance (in km)	0	10	20	30
Time (in minutes)	10 : 00	10 : 15	10 : 30	10 : 45

Answer: Mark time along the x-axis and distance along the y-axis, plot the four points and join them. The graph is a straight line, which shows that the car moves with a constant speed (10 km every 15 minutes, i.e. 40 km/h).

Consider two cars moving with velocity 10 m/s and 20 m/s respectively. Consider two situations: (i) When both cars are moving in same direction and (ii) When both cars are moving in opposite direction.

Answer: The question is incomplete in the source. It is meant to ask for the relative velocity of one car with respect to the other. (i) Same direction: relative velocity = 20 − 10 = 10 m/s (the faster car moves away from the slower one at 10 m/s). (ii) Opposite directions: relative velocity = 20 + 10 = 30 m/s (the cars approach or move apart at 30 m/s).

A 200 m train running with a speed of 50 m/s crosses a station, 300 m long. Find the time taken by the train to cross the station.

Answer: To cross the station the train must cover its own length plus the length of the station: 200 + 300 = 500 m. Time = distance ÷ speed = 500 ÷ 50 = 10 s.

A donkey is standing at a distance of 18 steps from a well. He moves towards the well in a manner that for every 5 steps taken in forward direction, he automatically moves back 3 steps. Now if his 1 step takes place in 1 sec, find the time after which he will fall into the well.

Answer: In each cycle (5 steps forward + 3 steps back = 8 steps = 8 s) the donkey gains 2 steps. After 7 cycles he has gained 14 steps (56 s). He is then 4 steps from the well. In the next cycle he moves 4 steps forward (4 s) and falls into the well before stepping back. Total time = 56 + 4 = 60 s.

A 100 m train is running at a speed of 15 m/s while another train which is 150 m is running at a speed of 10 m/s. Find the time taken by the train A to cross train B if

1. Both are moving in the same direction and engine of A is in alignment with the end of train B.
2. Both the trains are moving in opposite direction and engines of the train are in alignment with each other.

Answer: To cross completely, the trains must cover the sum of their lengths relative to each other: 100 + 150 = 250 m. (i) Same direction: relative speed = 15 − 10 = 5 m/s, so time = 250 ÷ 5 = 50 s. (ii) Opposite directions: relative speed = 15 + 10 = 25 m/s, so time = 250 ÷ 25 = 10 s.

Two bus stands A and B are located at a distance of 200 km. There is a 24 hour bus from each bus stand, at an interval of 10 minute. Now, if a bus starts from 8.00 A.M. from bus stand B, then find the total number of buses it will meet before reaching A, if each bus takes 2 hour to cover the distance of 200 km.

Answer: The bus from B takes 2 hours, so it reaches A at 10:00 A.M. It meets every bus from A that is on the road at some time between 8:00 and 10:00. Such buses left A between 6:00 A.M. (that bus arrives at B just as our bus leaves) and 10:00 A.M. (that bus leaves A just as our bus arrives). Buses leave A every 10 minutes, so from 6:00 to 10:00 there are 4 × 6 + 1 = 25 departures. Of these, the two at the ends (6:00 and 10:00) are met only at the bus stands, so the bus meets 23 buses on the way, and 25 buses if the two at the bus stands are counted.