NCERT Solutions for Class 7 Science Chapter 13 - Motion And Time

Answer:

Eyes do not blink after a fixed interval of time, so it is not a periodic phenomenon. Thus, blinking of eyes cannot be used for measurement of time.
Hence, the correct answer is option (d).

Answer:

Clock B has seconds hand, so it can measure seconds. But, clock A does not have seconds hand and hence seconds cannot be measured by clock A. Both the clocks have minute hand, so both can measure time interval of 5 minutes.
Hence, the correct answer is option (c).

Answer:

According to given data, given distance-time graph is true for both A and B.
Hence, the correct answer is option (a).

Answer:

Distance travelled = 54 km = 54000 m     (as, 1 km = 1000 m) 
Time = 90 min = 5400 s                 (as, 1 min = 60 s)
$$\begin{gathered} \mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{time}} \\ =\frac{54000}{5400} \\ =10 \mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

$$\begin{gathered} \mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}} \\ \Rightarrow \mathrm{S}=\frac{\mathrm{D}}{\mathrm{T}} \\ \Rightarrow \mathrm{S}\times \mathrm{T}=\mathrm{D} \end{gathered}$$
Hence, the correct answer is option (c).

Answer:

The time period of a simple pendulum is the time taken by it to travel from A to B and back to A.
Hence, the correct answer is option (a).

Answer:

Time period of a simple pendulum is the time taken by it to move from A to O, O to B and then back B to O & O to A. Time taken by pendulum to move from A to O is $\frac{1}{4}th$ of the time period of the pendulum.
Thus, Time period of simple pendulum = 4(Time taken by pendulum to move from A to O)
T = 4(t₁ + t₂)
Hence, the correct answer is option (d).

Answer:

$\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}$
SI unit of distance and time is  m and s respectively.
Hence, the SI unit of speed is m/s.
So, 5 m/s correctly represents speed.
Hence, the correct answer is option (a).

Answer:

$$\begin{gathered} \mathrm{Total} \mathrm{distance} =\left(3+3\right) \mathrm{km}=6 \mathrm{km} \\ \mathrm{Total} \mathrm{time}=\left(30+20\right) \min =50 \min =\frac{50}{60} \mathrm{h} \left(\mathrm{as}, 1 \mathrm{h}=60 \min \right) \\ \mathrm{Average} \mathrm{speed}=\frac{\mathrm{Total} \mathrm{distance}}{\mathrm{Total} \mathrm{time}} \\ =\frac{6 }{\left({\displaystyle \frac{50}{60}}\right)} \mathrm{km}/\mathrm{h}=\frac{60}{50}\times 6 \mathrm{km}/\mathrm{h} \\ =7.2 \mathrm{km}/\mathrm{h} \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

As speed of bob keeps on changing while oscillating between two points A and B, so its motion is said to be non-uniform.

Answer:

As $\mathrm{Speed}=\frac{\mathrm{distance}}{\mathrm{time}}$
So, their speeds will be different to cover different distances in the same time to reach the school.

Answer:

As we know that,
$Average speed=\frac{Total dis\tan ce covered}{Total time taken}$
So, speed of Boojho is higher than the speed of Paheli because Boojho takes less time ( 1 hour) as compared to Paheli who takes 2 hours to cover the same distance. 

Answer:

By analysing the given distance-time graph we can complete the table correctly as:
( It is important to note that the object travels equal distance in equal time interval)

Answer:

12 years and 3 months
 $$\begin{gathered} =(12\times 365 +3\times 30) \mathrm{days}=4470 \mathrm{days} \\ =(4470\times 24) \mathrm{hours} \\ =(4470\times 24\times 60) \mathrm{minutes} \\ =(4470\times 24\times 60\times 60) \mathrm{seconds} \\ =386,208,000 \mathrm{seconds} \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Speed}=\frac{\mathrm{distance}}{\mathrm{time}}=\frac{36000 \mathrm{km}}{1 \mathrm{hour}} \\ =\frac{36000 \mathrm{km}}{(60\times 60) \mathrm{s}} \left(\because 1 \mathrm{hour} = 60\times 60 \mathrm{s} \right) \\ =\frac{36000}{3600} \mathrm{km}/\mathrm{s} \\ =10 \mathrm{km}/\mathrm{s} \end{gathered}$$
 

Answer:

Answer:

Answer:

​(i) Position of the object at 20 s is 8 m.
(ii) Distance travelled by the object in 12 s is 6 m. 
(iii) Average speed = Total distance travelled/ total time taken $=\frac{8}{20} \mathrm{m}/\mathrm{s} =0.40 \mathrm{m}/\mathrm{s}$

Answer:

Total distance between Bholu's and Golu's house, d = 9 km 
Total time taken to cover this diatance, T = 1 hour
Time to cover remaining $\left(9-6=3 \mathrm{km}\right)$ distance,
 $$\begin{gathered} =60-(40+5)=15 \mathrm{minutes}=\frac{15}{60} \mathrm{h}=\frac{1}{4} \mathrm{h} \\ \mathrm{So}, \mathrm{speed} \mathrm{to} \mathrm{cover} \mathrm{second} \mathrm{half} \mathrm{of} \mathrm{the} \mathrm{journey} \\ =\frac{\mathrm{Remaining} \mathrm{distance}}{\mathrm{time} \mathrm{taken}}=\frac{3 \mathrm{km}}{\frac{1}{4} \mathrm{h}} \\ =12 \mathrm{km}/\mathrm{h} \\ \mathrm{Now} \mathrm{average} \mathrm{speed} \mathrm{for} \mathrm{the} \mathrm{entire} \mathrm{journey}, \\ =\frac{\mathrm{Total} \mathrm{distance}}{\mathrm{Total} \mathrm{time} \mathrm{taken}}=\frac{d}{T}=\frac{9 \mathrm{km}}{1 \mathrm{h}} \\ =9 \mathrm{km}/\mathrm{h} \end{gathered}$$
 

Answer:

(a) As the position of  Boojho remains unchanged between  B and C, this indicates that he is at rest (speed = 0) in the time interval 4 to 8 minutes.
(b) The motion of Boojho between 0 to 4 minutes is non-uniform. 
(c)  The speed between 8 and 12 minutes of his journey:
 $$\begin{gathered} \mathrm{Speed}=\frac{\mathrm{distance} \mathrm{covered}}{\mathrm{time} \mathrm{to} \mathrm{cover} \mathrm{the} \mathrm{distance}}=\frac{\left(225-150\right) \mathrm{m}}{\left(12-8\right) \mathrm{minutes}} \\ =\frac{75}{4} \mathrm{m}/\mathrm{minute}=18.75 \mathrm{m}/\mathrm{minute} \end{gathered}$$