CBSE Class 7 Science Chapter 13: Motion and Time NCERT Solutions

The different types of motion can be classified as follows:

(i) Motion of your hands while running: This is an example of oscillatory motion, as your hands move back and forth rhythmically.

(ii) Motion of a horse pulling a cart on a straight road: This is an example of motion along a straight line, as both the horse and the cart move in a straight direction.

(iii) Motion of a child in a merry-go-round: This is an example of circular motion, as the child moves in a circular path.

(iv) Motion of a child on a see-saw: This is an example of oscillatory motion, as the child moves up and down repeatedly around a fixed point.

(v) Motion of the hammer of an electric bell: This is an example of oscillatory motion, as the hammer strikes the bell repeatedly, moving back and forth.

(vi) Motion of a train on a straight bridge: This is an example of motion along a straight line, as the train moves forward on a straight track.

Let's evaluate each statement:

(i) The basic unit of time is second - Correct. The second is the standard SI unit for time.

(ii) Every object moves with a constant speed - Not correct. Many objects change their speed during their motion. For example, a car speeds up or slows down.

(iii) Distances between two cities are measured in kilometers - Correct. Kilometers are a suitable unit for measuring large distances like those between cities.

(iv) The time period of a given pendulum is not constant - Not correct. For a simple pendulum, the time period is constant for small amplitudes of swing, although it can vary slightly with factors like length and gravity. The statement as presented is often considered incorrect in introductory physics contexts where it's assumed constant.

(v) The speed of a train is expressed in m/h - Not correct. The speed of a train is typically expressed in kilometers per hour (km/h) or meters per second (m/s), not meters per hour (m/h).

The time period of a pendulum is defined as the time taken to complete one full oscillation. We can calculate it using the given information:

Total time taken = 32 seconds

Number of oscillations = 20

The formula for time period is:

$$Time\backslash \; period\; =\; \backslash frac\{Total\backslash \; time\backslash \; taken\}\{Number\backslash \; of\backslash \; oscillations\}$$

Substituting the values:

$$Time\backslash \; period\; =\; \backslash frac\{32\backslash \; s\}\{20\backslash \; oscillations\}\; =\; 1.6\backslash \; seconds$$

Therefore, the time period of the pendulum is 1.6 seconds.

We are given the distance and the time taken by the train. We can calculate the speed using the formula:

$$Speed\; =\; \backslash frac\{Distance\backslash \; covered\}\{Time\backslash \; taken\}$$

Given:

Distance = 240 km

Time = 4 hours

Substituting these values into the formula:

$$Speed\; =\; \backslash frac\{240\backslash \; km\}\{4\backslash \; h\}\; =\; 60\backslash \; km/h$$

The speed of the train is 60 km/h.

First, let's find the distance moved by the car and the time taken.

Initial odometer reading = 57321.0 km

Final odometer reading = 57336.0 km

Distance moved by car = Final reading - Initial reading

$$Distance\; =\; 57336.0\backslash \; km\; -\; 57321.0\backslash \; km\; =\; 15.0\backslash \; km$$

Now, let's find the time taken:

Start time = 08:30 AM

End time = 08:50 AM

Time taken = 08:50 AM - 08:30 AM = 20 minutes

To calculate speed in km/min:

$$Speed\; =\; \backslash frac\{Distance\backslash \; covered\}\{Time\backslash \; taken\}\; =\; \backslash frac\{15\backslash \; km\}\{20\backslash \; min\}\; =\; 0.75\backslash \; km/min$$

Next, let's express the speed in km/h. First, convert the time taken to hours:

Time taken = 20 minutes = $\backslash frac\{20\}\{60\}$ hours = $\backslash frac\{1\}\{3\}$ hour

Now, calculate the speed in km/h:

$$Speed\; =\; \backslash frac\{Distance\backslash \; covered\}\{Time\backslash \; taken\}\; =\; \backslash frac\{15\backslash \; km\}\{\backslash frac\{1\}\{3\}\backslash \; h\}\; =\; 15\; \backslash times\; 3\backslash \; km/h\; =\; 45\backslash \; km/h$$

So, the speed of the car is 0.75 km/min, which is equal to 45 km/h.

We are given the speed of the bicycle and the time Salma takes to reach school. We need to calculate the distance.

Speed = 2 m/s

Time taken = 15 minutes

To use the formula Distance = Speed × Time, the units must be consistent. Let's convert the time from minutes to seconds:

$$Time\; =\; 15\backslash \; minutes\; \backslash times\; 60\backslash \; seconds/minute\; =\; 900\backslash \; seconds$$

Now, we can calculate the distance:

$$Distance\; =\; Speed\; \backslash times\; Time\; =\; 2\backslash \; m/s\; \backslash times\; 900\backslash \; s\; =\; 1800\backslash \; meters$$

The distance is 1800 meters. To express this in kilometers, we divide by 1000:

$$Distance\; =\; \backslash frac\{1800\}\{1000\}\backslash \; km\; =\; 1.8\backslash \; km$$

The distance between Salma's house and the school is 1.8 km.

A distance-time graph plots the distance covered by an object on the y-axis against the time taken on the x-axis.

1. A car moving with a constant speed:

When an object moves with a constant speed, it covers equal distances in equal intervals of time. On a distance-time graph, this is represented by a straight line sloping upwards from the origin. The slope of this line represents the constant speed.

(Imagine a graph with 'Time' on the horizontal axis and 'Distance' on the vertical axis. A straight line starts from the bottom-left corner (origin) and goes up towards the top-right.)

2. A car parked on a side road:

When a car is parked, its distance from a reference point does not change over time. On a distance-time graph, this is represented by a horizontal line parallel to the time axis. This indicates that the distance remains constant, meaning the object is stationary.

(Imagine a graph with 'Time' on the horizontal axis and 'Distance' on the vertical axis. A horizontal line is drawn at a certain distance value, showing that as time increases, the distance stays the same.)