**Lesson: **Motion and Time

Which of the following cannot be used for measurement of time?

(a) A leaking tap.

(b) Simple pendulum.

(c) Shadow of an object during the day.

(d) Blinking of eyes.

(d) Blinking of eyes

Time interval between two events can be measured in other options. But in case of blinking of eyes, it cannot be measured as it does not happen at the same rate.

Two clocks A and B are shown in the figure below. Clock A has an hour and a minute hand, whereas clock B has an hour hand, minute hand as well as a second hand. Which of the following statement is correct for these clocks?

(a) A time interval of 30 seconds can be measured by clock A.

(b) A time interval of 30 seconds cannot be measured by clock B.

(c) Time interval of 5 minutes can be measured by both A and B.

(d) Time interval of 4 minutes 10 seconds can be measured by clock A.

(c) Time interval of 5 minutes can be measured by both A and B as both the clocks have the minute hand which can measure time intervals by minutes.

Two students were asked to plot a distance time graph for the motion described by Table A and Table B.

The graph given in the figure above is true for

(a) Both A and B.

(b) A only.

(c) B only.

(d) Neither A and B.

(a) Both A and B

A bus travels
54 km in 90 minutes*. *The speed of the bus is

(a) 0.6 m/s

(b) 10 m/s

(c) 5.4 m/s

(d) 3.6 m/s

(b) 10 m/s

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

$=54\times \frac{60}{90}$

=36 km/h

$\text{Speedin}m/s=36\times \frac{5}{18}$

=10 m/s

If we denote speed by S, distance by D and time by T, the relationship between these quantities is

(a) S = DT

(b) $\text{T}=\frac{\text{S}}{\text{D}}$

(c) $S=\frac{1}{\text{T}}\times \text{D}$

(d) $\text{S=}\frac{\text{T}}{\text{D}}$

(c) $\text{S=}\frac{\text{1}}{\text{T}}\text{\xd7D}$

Observe the figure below.

The time period of a simple pendulum is the time taken by it to travel from

(a) A to B and back to A.

(b) O to A, A to B and B to A.

(c) B to A, A to B and B to O.

(d) A to B.

(a) A to B and back to A.

Time period is the time taken to complete one oscillation.

The figure below shows an oscillating pendulum.

Time taken by
the bob to move from A to C is *t*_{1} and from C to O is *t*_{2}.
The time period of this simple pendulum is

(a) (t_{1} + t_{2})

(b) 2(t_{1} + t_{2})

(c) 3(t_{1} + t_{2})

(d) 4(t_{1} + t_{2})

(d) 4(t_{1}
+ t_{2})

The correct symbol to represent the speed of an object is

(a) 5 m/s

(b) 5 mp

(c) 5 m/s^{-1}

(d) 5 s/m

(a) 5 m/s

Boojho walks to his school which is at a distance of 3 km from his home in 30 minutes. On reaching he finds that the school is closed and comes back by a bicycle with his friend and reaches home in 20 minutes. His average speed in km/h is

(a) 8.3

(b) 7.2

(c) 5

(d) 3.6

(b) 7.2

$Averagespeed=\frac{total\text{\hspace{0.17em}}distance\text{\hspace{0.17em}}travelled}{total\text{\hspace{0.17em}}time\text{\hspace{0.17em}}taken}$

Average speed $=\frac{\left(\text{3km}+\text{3km}\text{}\right)}{\left(30min+\text{20min}\right)}\text{}$

Average speed $=\frac{\text{6km}}{\text{50min}}\text{}$

$=\text{}\text{6km}\times \frac{\text{60}}{\text{50hr}}$

= 7.2 km/h

A simple pendulum is oscillating between two points A and B as shown in the figure below. Is the motion of the bob uniform or non-uniform?

The motion of the bob is non-uniform.

Paheli and Boojho have to cover different distances to reach their school but they take the same time to reach the school. What can you say about their speed?

Since they cover different distances in the same interval of time, their speed is different.

If Boojho covers a certain distance in one hour and Paheli covers the same distance in two hours, who travels in a higher speed?

Boojho travels with higher speed because he takes lesser time than Paheli to cover the same distance.

Complete the data of the table given below with the help of the distance-time graph given in the figure below.

Distance (m) |
0 |
4 |
8 |
12 |
16 |
20 |

Time (s) |
0 |
2 |
4 |
6 |
8 |
10 |

The average age of children of Class VII is 12 years and 3 months. Express this age in seconds.

1 year = 365 days

12 years 3 months $=12\times 365+3\times 30$

= 4470 days

$1\text{day}=24\text{hours}\times 60\text{minutes}\times 60\text{seconds}$

= 86400 second

$4470\text{days}=4470\times 86400$

**= **386208000
second

A spaceship travels 36,000 km in one hour. Express its speed in km/s.

$\frac{\text{36000km}}{1\text{\hspace{0.17em}}\text{hr}}$

$=\frac{\text{36000km}}{3600\text{\hspace{0.17em}}\text{s}}$

= 10 km/s

Starting from A, Paheli moves along a rectangular path ABCD as shown in the figure below. She takes 2 minutes to travel each side. Plot a distance-time graph and explain whether the motion is uniform or non-uniform.

From the graph it is clear that Paheli covers unequal distances in equal interval of time. Therefore, the motion is non-uniform.

Plot a
distance-time graph of the tip of the second hand of a clock by selecting 4
points on *x-*axis and *y-*axis respectively. The circumference of
the circle traced by the second hand is 64 cm.

Since the circumference traced by the second hand is 64 cm, we can consider the distance covered by the tip of the second hand to be 64 cm per minute or per 60 seconds.

Accordingly, we can form the table as given below.

Given below as figure is the distance-time graph of the motion an object.

(i) What will be the position of the object at 20s?

(ii) What will be the distance travelled by the object in 12s?

(iii) What is the average speed of the object?

(i) 8 m from the starting point

(ii) 6 m from the starting point

(iii) Average speed = $\frac{\text{Totaldistancetravelled}}{\text{totaltimetaken}}$

$=\frac{8m}{20s}$

= 0.4 m/s

Distance between Bholu’s and Golu’s house is 9 km. Bholu has to attend Golu’s birthday party at 7 o’clock. He started from his home at 6 o’clock on his bicycle and covered a distance of 6 km in 40 minutes. At that point he met Chintu and he spoke to him for 5 minutes and reached Golu’s birthday party at 7 o’clock. With what speed did he cover the second part of the journey? Calculate his average speed for the entire journey.

Distance covered in the second part of the journey =

total distance - distance covered in the first part of the journey

= 9 km - 6 km

= 3 km

Time taken to cover the distance = total time $\u2013$ time taken to cover the first part of the journey $\u2013$ time spent

Time taken to cover the distance = 60 min $\u2013$ 40 min $\u2013$ 5 min = 15 min

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

$=\frac{3km}{\frac{15}{60}hr}$

$=\frac{3\times 60}{15}km/h$

$=12\text{\hspace{0.17em}}km/h$

Average speed = total distance travelled/ total time taken

Average speed $=\frac{9\text{\hspace{0.17em}}\text{km}}{\frac{55}{60}\text{hr}}$

$=\frac{9\times 60}{55}\text{km/h}$

Average speed = 9.81 km/h

Boojho goes to the football ground to play football. The distance-time graph of his journey from his home to the ground is given in the figure below.

(a) What does the graph between point B and C indicate about the motion of Boojho?

(b) Is the motion between 0 to 4 minutes uniform or non-uniform?

(c) What is his speed between 8 and 12 minutes of his journey?

(a) The graph between points B and C indicates that Boojho is at rest and his speed is zero.

(b) From the graph we see that unequal distance is covered in equal interval of time. Therefore, the motion is non-uniform.

(c) At the 8th minute Boojho has covered 150 m and at the 12th minute Boojho has covered 225 m.

The distance
covered by Boojho between the 8^{th }and 12^{th} min

Time taken to cover 75 m = 12 - 8

= 4 min

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

$=\frac{75}{4}$

= 18.75 m/min