Motion And Time

The activity can be performed as follows:

1. Find the latitude of the city with the help of Atlas.

2. Cut out a triangular piece of a cardboard such that its one angle is equal to the latitude of the location and the angle opposite to it is a right angle.

3. Fix this piece, called gnomon, vertically along a diameter of a circular board by making a groove along a diameter on the circular board.

4. Choose a space, which receives sunlight for most of the day. Mark a line on the ground along the North-South direction. Place the sundial in the sun.

5. Mark the position of the tip of the shadow of the gnomon on the circular board as early in the day as possible. Mark the position of the tip of the shadow every hour throughout the day. Draw lines to connect each point marked by you with the centre of the base of the gnomon.

6. Extend the lines on the circular board up to its periphery.

This Sundial can be used for reading the time at chosen place.

(i) Motion of hands while running are in oscillatory motion. They moves in a to-and fro motion and completes an oscillation when the hands moves from one extreme end to another extreme end and returns back to initial position. This type of motion is known as oscillatory motion.

(ii)Motion of a horse pulling a cart on a straight road is a straight-line motion because horse is pulling the cart on a straight line so the motion will be in a single direction and in straight line.

(iii) Motion of a child in a merry-go-round is circular motion because merry-go-round swing tends to move in a circular path and comes back to initial position after each round.

(iv) Motion of a child on a see-saw is an oscillatory motion. A see-saw moves in to-and fro- motion completes an oscillation when it moves from one extreme end to another extreme end and returns back to initial position.

(v) Motion of the hammer on an electric bell is also an oscillatory motion, because the hammer hits the bell and it vibrates rapidly.

(vi) Motion of a train on a straight bridge is a straight-line motion because on a straight bridge, the train will move straight as according the bridge.

Every object moves with a constant speed. This is not correct because some objects can move with changing speed. This type of motion is known as non-uniform motion.

The time period of a given pendulum is not constant. This statement is not correct because moves in a to-and-fro motion and completes an oscillation when the bob moves from one extreme end to another extreme end and returns back to the initial position. This type of motion is known as oscillatory motion. The time taken by the pendulum to complete one oscillation is known as the time period of a given pendulum. The time period doesn’t vary with a slight change in its initial displacement and hence the time period remains constant.

The speed of the train is expressed in m/h. This statement is not correct. The speed of a train is expressed in km/h.

The time measuring devices that were used in the ancient times in different parts of the world are:

1. Sun Dial

2. Water clocks

3. Sand clocks.

Sun Dial:

This time measuring device uses a spot of light or shadow by the sun’s position on a reference scale. The prototypes of Sundials are seems to be originated in Russia and they were used in 1830s.

Water Clocks:

This time measuring device uses regulated water flow of liquid i.e. inflow or outflow of liquid to calculate the time. They are supposed to be existed in 16^th Century in Egypt.

Sand Clocks:

The sand clocks are supposed to be invented at Alexandria about 150 B.C.

No. of oscillations, n = 20

Time taken to complete 20 oscillations, t= 32 seconds

Time period of the pendulum i.e. time taken for 1 oscillation, T =

Time period of pendulum =


=1.6 seconds.

So, the time period of pendulum is 1.6 seconds.

1. Take one plastic cup and place it in upside down on a table. Put the other cup on the top of the first one so that the bottom parts of both cups touch each other.

2. Place two lids on the cups and make holes into them. And then invert one cup one another.

3. Measure the sand and put on top of one cup timer. Allow the sand to pass through the holes.

4. Place the cup timer on top of plate and note down the time taken by the sand to pass through the bottom of the timer.

5. Adjust the amount of sand until the time taken by it is 2 minutes for all the sand to pass through the bottom of the timer.

The distance between two stations, d = 240 km

Time taken by the train to cover the distance, t = 4 h

Speed of the train =

Speed =


=60 km/h.

1. Oscillate the swing without any person sitting on it and note down the time period for 5 oscillations.

2. Ask a person to sit on the swing and note down the time for 5 oscillations.

3. Repeat the same for different persons.

4. Calculate the average time taken by dividing to no. of oscillations.

The time period of the swing will be different for different cases. The time period of swing when no one is sitting on it will be least and will depend on the weight of the object sitting on the swing. The higher is the mass of the object on swing, the more will be the time taken by the swing to complete one oscillation because speed becomes less.

when time is 08:30 a.m.: the distance read by Odometer = 5,7321.0 km

When time is 08:50 a.m.: The distance read by Odometer = 57,336.0 km

Distance traveled by car = distance read by odometer at 08:50 am – distance read by Odometer at 08:30 am

= 57,336.0 km- 5,7321.0 km

Distance travelled= 15 km

Time taken = Final time – initial time = 8:50 am – 8:30 am

Time taken t = 20 minutes

Speed of the car=

=

Converting minutes into hours, divide by 60 as 1 hour = 60 minutes

So, 1 minutes= h

So, speed = 15 x

Speed = 45 km/h.

Speed of bicycle = 2m/s

Time taken by Salma to reach her school = 15 min

Converting into seconds, t= 15 x 60 = 900 seconds.

Distance between her house and the school = speed x time

Distance = 2 x 900

Distance = 1800 m

In km distance is given by = 1800/1000

Distance between her house and school = 1.8 km.

(i) Since a car is moving with a constant speed, so the motion will be a uniform motion and the time will vary proportionally with distance. The distance time graph for motion of an object with constant speed is a straight line. So, the graph of distance time for a car moving with a constant speed will be as follows:

(ii) Since a car parked on a road side, it will not be in a motion and therefore there will be no variation in distance.

The correct expression is: Speed=

_{Once, we know the distance and time taken to travel the distance, the speed of any object can be calculated.}

The basic unit to measure time is second and for distance, it is meter. So, the basic unit to measure speed is m/s.

Speed of the car in the first 15 minutes = 40 km/h

So, distance traveled in the first 15 minutes will be = speed x time

Distance = 40km/h x 15 minutes

Distance = (40 x 15/60) km = 10 km

Speed in next 15 minutes = 60 km/h

Distance traveled in next 15 min= speed in next 15 minutes × time

= 60 km/h x 15 min

= h

Distance in next 15 minutes = 15 km

The total distance covered by car = distance in first 15 min+ distance in next 15 min

Total distance = 10+15


= 25 km.

Distance travelled by the blue car = 2 cm

And 1cm = 100 m (as given in the question)

2 cm = 200 m

Therefore, distance travelled by blue car = 200 m

Time taken = 10 sec

Speed of the blue car = distance/ time taken = 200/10s

Speed of the blue car = 20m/s

Vehicle A is moving faster than vehicle B as according to the graph given in the above figure, the slope of graph A is more than slope of graph B. It is clear from the above graph. That more distance has been travelled by the vehicle A in lesser time and therefore, the speed will be more than vehicle B.

In option (iii), the distance time graph is not a straight line, instead it is a curve. The distance time graph for an object moving with a constant speed is straight line. This means that the speed is not constant is changing. So, the distance-time graph of option (iii) shows a truck moving with speed which is not constant.