**Lesson: **Motion and time

Classify the following as motion along a straight line, circular or oscillatory motion:

i) Motion 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.

i) Motion of your hands while running. - Oscillatory motion

ii) Motion of a horse pulling a cart on a straight road. - Straight line motion

iii) Motion of a child in a merry-go-round. - Circular motion

iv) Motion of a child on a see-saw. - Oscillatory motion

v) Motion of the hammer of an electric bell. - Oscillatory motion

vi) Motion of a train on a straight bridge. - Straight line motion

Which of the following are not correct?

(i) The basic unit of time is second.

(ii) Every object moves with a constant speed.

(iii) Distances between two cities are measured in kilometers.

(iv) The time period of a given pendulum is not constant.

(v) The speed of a train is expressed in m/h.

(i) True

(ii) False. Different objects have different speed.

(iii) True

(iv) False. The time period of a given pendulum is fixed.

(v) False. The speed of a train is expressed in km/h.

A simple pendulum takes 32 s to complete 20 oscillations. What is the time period of the pendulum?

No. of oscillations = 20

Time taken for 20 oscillations = 32

Time period $=\frac{Total\text{\hspace{0.17em}}time\text{\hspace{0.17em}}taken}{No.\text{\hspace{0.17em}}of\text{\hspace{0.17em}}oscillations}$

Time period $=\frac{32}{20}$

= 1.6 s

The distance between two stations is 240 km. A train takes 4 hours to cover this distance. Calculate the speed of the train.

Distance between two stations is 240 km

Time taken to cover the distance is 4 hours

$Speed=\frac{Distance}{Time\text{\hspace{0.17em}}taken}$

$=\frac{240}{4}$

= 60 km/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 km/min during this time. Express the speed in km/h also.

Odometer reading of car at 08:30 a.m. = 57321.0 km

Odometer reading of car at 08:50 a.m. = 57336.0 km

Total distance travelled by the car

= 57336.0 - 57321.0

Total distance travelled by the car = 15 km

Time taken to cover this distance

= 08:50 a.m. - 08:30 a.m. = 20 min

Speed of the car in km/min = $\frac{Distance}{Time\text{\hspace{0.17em}}taken}$

$=\frac{15}{20}$

= 0.75 km/h

Speed of the car in km/hr = $\frac{Distance}{Time\text{\hspace{0.17em}}taken}$

$=\frac{15\times 60}{20}$

$=45\text{}\text{km/hr}\text{}$

Salma takes 15 minutes from her house to reach her school on a bicycle. If the bicycle has a speed of 2 m/s, calculate the distance between her house and the school.

Speed of Salma’s bicycle = 2 m/s

Time taken by Salma to reach her school

= 15 min

$\begin{array}{l}=15\times 60\\ =900\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{second}\end{array}$

Distance = Speed × time

Distance

$=2\times 900$

= 1800 m

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.

Which of the following relations is correct?

(i) $\text{Speed}=\text{Distance}\times \text{Time}$

(ii) $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

(iii) $\text{Speed}=\frac{\text{Time}}{\text{Distance}}$

(iv) $\text{Speed}=\frac{1}{\text{Distance}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\times \text{Time}}$

(ii)
$\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

The basic unit of speed is:

(i) km/min

(ii) m/min

(iii) km/h

(iv) m/s

(iv) m/s

A car moves with a speed of 40 km/h for 15 minutes and then with a speed of 60 km/h for the next 15 minutes. The total distance covered by the car is:

(i) 100 km

(ii) 25 km

(iii) 15 km

(iv) 10 km

(ii) 25 km

$\text{Distance}=\text{Speed}\times \text{Time}$

Distance covered in first $15\text{min}$$=40\text{km/h}\text{}\times \frac{15}{60}$

=10.00 km

Distance covered in next 15 min

$=60\text{}\text{km/h}\text{}\times \frac{15}{60}$

= 15.00 km

Total distance covered

$=\frac{10}{15}$

= 25 km

Suppose the two photographs, shown in the figure A and figure B, had been taken at an interval of 10 seconds. If a distance of 100 meters is shown by 1 cm in these photographs, calculate the speed of the blue car.

Distance moved by the blue car in 10 seconds = 2 cm

Given 1 cm =100 m

Therefore, Distance moved by the blue car in 10 seconds = 200 m

$\text{Speed}=\frac{\text{Distance}}{\text{time}}$

$=\frac{200}{10}$

= 20 m/s

The figure below shows the distance-time graph for the motion of two vehicles A and B. Which one of them is moving faster?

Figure: Distance-time graph for the motion of two cars

From the figure, the slope of vehicle A is steeper than slope of vehicle B.

The steeper the line, greater the speed. Therefore vehicle A is moving faster.

Which of the following distance-time
graphs shows a truck moving with speed which is not constant?

(i)