CBSE Class 7 Science Chapter 13 Motion and Time NCERT Solutions

The classification of the given motions is as follows:

1. Motion of your hands while running: This involves a back-and-forth movement around a central point, hence it is oscillatory motion.
2. Motion of a horse pulling a cart on a straight road: The horse and cart are moving along a straight path, so this is motion along a straight line.
3. Motion of a child in a merry-go-round: The child moves in a circular path, hence this is circular motion.
4. Motion of a child on a see-saw: A see-saw moves up and down repeatedly around a pivot point, which is characteristic of oscillatory motion.
5. Motion of the hammer of an electric bell: The hammer strikes the bell and then returns, repeating this action. This to-and-fro movement is oscillatory motion.
6. Motion of a train on a straight bridge: The train moves along a straight track, hence this is motion along a straight line.

Let's evaluate each statement:

1. The basic unit of time is second: Correct. The second is the standard SI unit for time.
2. Every object moves with a constant speed: Not correct. Many objects change their speed during motion; constant speed means the speed does not change over time.
3. Distances between two cities are measured in kilometers: Correct. Kilometers are a suitable unit for measuring large distances like those between cities.
4. The time period of a given pendulum is not constant: Not correct. For a simple pendulum, the time period is generally constant under specific conditions (like small amplitude swings).
5. 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 sometimes meters per second (m/s), but not meters per hour (m/h).

Therefore, the incorrect statements are (ii), (iv), and (v).

The time period of a pendulum is defined as the time it takes 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:

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

Thus, 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

Calculating the speed:

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

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

The distance moved by the car is the difference between the final and initial readings:

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

The time interval is from 08:30 AM to 08:50 AM.

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

To calculate the speed in km/min:

$$Speed\; =\; \backslash frac\{Distance\backslash \; covered\}\{Time\backslash \; taken\}\; =\; \backslash frac\{15.0\backslash \; km\}\{20\backslash \; min\}\; =\; 0.75\backslash \; km/min$$

Now, let's express the speed in km/h. We need to convert the time taken into hours.

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

Calculating the speed in km/h:

$$Speed\; =\; \backslash frac\{Distance\backslash \; covered\}\{Time\backslash \; taken\}\; =\; \backslash frac\{15.0\backslash \; km\}\{\backslash frac\{1\}\{3\}\backslash \; h\}\; =\; 15.0\; \backslash times\; 3\backslash \; km/h\; =\; 45\backslash \; km/h$$

The distance moved by the car is 15.0 km. The speed of the car is 0.75 km/min, which is equal to 45 km/h.

We are given Salma's speed and the time she 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. It is often useful to express this distance in kilometers:

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

The distance between Salma's house and the school is 1800 meters or 1.8 km.

A distance-time graph plots the distance traveled by an object against the time elapsed. The shape of the graph provides information about the object's motion.

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 x-axis and Distance on the y-axis. The line starts at (0,0) and goes up diagonally to the 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 x-axis and Distance on the y-axis. The line is horizontal at a certain distance value, showing no change in distance as time progresses.)