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Page No 184:
Question 1:
Which of these activities can be measured in hours?
(a) writing one page
(b) eating a sandwich
(c) a journey from Chennai to Mumbai
(d) watching a TV serial
Answer:
(c) A journey from Chennai to Mumbai
Writing a page and eating a sandwich need a smaller unit than hour. Watching a TV serial can be measured in hours, but the most appropriate option is the journey from Chennai to Mumbai.
Page No 184:
Question 2:
The speed of a moving object means
(a) how far it goes.
(b) how long it goes on moving.
(c) how far it goes in a certain time.
(d) none of these.
Answer:
(c) how far it goes in a certain time
Speed is the rate at which an object covers a given distance in a particular period of time.
Page No 184:
Question 3:
On Saturday night, Prachi spent 18 minutes on her social science homework, 35 minutes on her mathematics homework and 22 minutes on her English homework. How much time did she spend on her homework in total?
(a) 1 h 75 min
(b) 2 h 15 min
(c) 2h 75 min
(d) 1 h 15 min
Answer:
(d) 1 h 15 min
Time spent by Prachi on social science homework = 18 min
Time spent by Prachi on mathematics homework = 35 min
Time spent by Prachi on English homework = 22 min
Total time spent by Prachi on her homework = 18 + 35 + 22
=75 min or (60 + 15) min (∵ 1 h = 60 min)
= 1 h 15 min
Page No 184:
Question 4:
If a moving body covers equal distances in equal intervals of time, it is said
(a) to have the same velocity.
(b) to have uniform motion.
(c) to have non-uniform motion
(d) none of the above
Answer:
(b) to have uniform motion
Uniform motion is a motion in which a body covers equal distances in equal intervals of time. This suggests that the body moves at a constant speed.
Page No 184:
Question 5:
Study the table given below. Tick the correct statement.
Distance Travelled | Time Taken | |
Vikram | 144 km | 2 h |
Tarun | 25 m | 2 s |
Ram | 100 m | 6 s |
(a) Vikram travels fastest.
(b) All three travel at the same speed.
(c) Tarun and Ram travel at the same speed.
(d) Ram travels at the greatest speed.
Answer:
(a) Vikram travels the fastest.
Distance travelled by Vikram = 144 km
= (144 $\times $ 1000) m
= 144000 m (∵ 1 km = 1000 m)
Time taken by him = 2 h
= (2 $\times $ 60 $\times $ 60) s
= 7200 s (∵ 1 h = 60 $\times $ 60 s)
Speed of Vikram = Distance travelled/Time taken
= 144000/7200
= 20 m/s
Similarly,
Speed of Tarun = Distance travelled by Tarun/Time taken
= 25/2
= 12.5 m/s
And,
Speed of Ram = Distance travelled by Ram/Time taken
= 100/6
= 16.7 m/s
It is clear from the above calculation that Vikram travels the fastest among the three.
Page No 184:
Question 6:
A. A train runs from New Delhi to Kolkata. It first covers a distance of 400 km in 7 hours and then a distance of 550 km in 8 hours.
B. Ratna takes part in a car race. She drivers a distance of 70 km in the first, second and third hours.
(a) A is an example of uniform motion and B is an example of non-uniform motion.
(b) A is an example of non-uniform motion and B is an example of uniform motion.
(c) A and B are examples of uniform motion.
(d) A and B are example of non-uniform motion.
Answer:
(b) A is an example of non-uniform motion and B is an example of uniform motion.
A. Speed of the train in the first part of the journey = 400/7
= 57.14 km/h
Speed of the train in the second part of the journey= 550/8
= 68.75 km/h (∵ Speed = Distance/Time)
In both the parts, the train has different speeds. Hence, this is an example of non-uniform motion.
B. Ratna travels a distance of 70 km in the first, second and third hours. It means that she is travelling equal distances in equal intervals of time. In other words, she is travelling at a constant speed. Thus, this is an example of uniform motion.
Page No 184:
Question 7:
This graph shows−
(a) the motion of your school bus as it picks up students.
(b) the motion of Ravi, who stops at the market on his way back home from school.
(c) the motion of an ant as it collects rice grains.
(d) the motion of an athlete running a 200 m race.
Graph
Answer:
(b) the motion of Ravi, who stops at the market on his way back home from the school.
Ravi stops for (11 $-$ 7) minutes, that is, 4 minutes.
Page No 185:
Question 8:
In which of these graphs is the object at rest?
(a) graph 1
(b) graph 2
(c) graph 3
(d) graph 4
Answer:
(b) Graph 2
Here, the distance of the object is constant with respect to time.
Page No 185:
Question 9:
In which of these graphs is the speed of the moving object constant?
(a) graph 1
(b) graph 2
(c) graph 3
(d) graph 4
Answer:
(a) Graph 1
The distance–time graph of an object moving at a constant speed is always represented by a straight line.
Page No 185:
Question 10:
The graph shows the motion of two vehicles A and B. Which one of them is moving with greater speed?
(a) A
(b) B
(c) both A and B are moving with the same speed.
(d) Cannot say from these graphs.
graph
Answer:
(a) A
We know:
Speed = Distance/Time
The above relation shows that the speed of a vehicle is greater if it covers maximum distance in a given interval of time. To compare the distances, draw a line perpendicular to the time axis.
It is now evident that for a given time t, the distance covered by vehicle A is more than that covered by vehicle B. Hence, vehicle A is moving with a greater speed.
Page No 185:
Question 11:
Express these times according to the 24-hour clock.
(a) 2.25 am
(b) 3.05 am
(c) 10.50 pm
(d) 8.05 pm
(e) 7.05 pm
Answer:
(a) 2:25 a.m.
This is 2 hours after the beginning of a day, so the 24-hour clock will show the time as 0225.
(b) 3:05 a.m.
This is 3 hours after the beginning of a day, so the 24-hour clock will show the time as 0305.
(c) 10:50 p.m.
This is 22 hours (12 + 10) after the beginning of a day, so the 24-hour clock will show the time as 2250.
(d) 8:05 p.m.
This is 20 hours (12 + 8) after the beginning of a day, so the 24-hour clock will show the time as 2005.
(e) 7:05 p.m.
This is 19 hours (12 + 7) after the beginning of a day, so the 24-hour clock will show the time as 1905.
Page No 185:
Question 12:
Express these times according to the 12-hour clock.
(a) 0050 hr
(b) 1650 hr
(c) 1830 hr
(d) 1007 hr
(e) 2345 hr
Answer:
(a) 12:50 a.m.
(b) 4:50 p.m.
(c) 6:30 p.m.
(d) 10:07 a.m.
(e) 11:45 p.m.
Page No 185:
Question 13:
Express the following speeds in m/s.
(a) 45km/h
(b) 135 km/h
(c) 90 km/h
(d) 75km/h
Answer:
We know:
1 km = 1000 m
1 h = (60 $\times $ 60 ) s = 3600 s
Now,
(a) 45 km/h = (45 $\times $ 1000)/(1 $\times $ 3600)
= 12.5 m/s
(b) 135 km/h = (135 $\times $ 1000)/(1 $\times $ 3600)
= 37.5 m/s
(c) 90 km/h = (90 $\times $ 1000)/(1 $\times $ 3600)
= 25 m/s
(d) 75 km/h = (75 $\times $ 1000)/(1 $\times $ 3600)
= 20.83 m/s
Page No 185:
Question 14:
Express the following speeds in km/h.
(a) 65km/h
(b) 40km/h
(c) 100m/s
(d) 10 m/s
Answer:
We know:
1000 m = 1 km
1 m = (1/1000) km
Also,
(60 $\times $ 60) s or 3600 s = 1 h
1 s = (1/3600) h
(a) 65 m/s $=\frac{(65/1000)}{(1/3600)}$
$=234\mathrm{km}/\mathrm{h}$
(b) 40 m/s $=\frac{(40/1000)}{(1/3600)}$
$=144\mathrm{km}/\mathrm{h}$
(c) 100 m/s $=\frac{(100/1000)}{(1/3600)}$
$=360\mathrm{km}/\mathrm{h}$
(d) 10 m /s $=\frac{(10/1000)}{(1/3600)}$
$=36\mathrm{km}/\mathrm{h}$
Page No 185:
Question 15:
An athlete covers 1500 m in 4 minutes. Calculate his speed in m/s and km/h.
Answer:
We know:
1 km = 1000 m
And,
1 h = 60 min
= (60 $\times $ 60) s
= 3600 s
Now,
Distance travelled by the athlete = 1500 m
= (1500/1000) km
= 1.5 km
Time taken = 4 min
= 4/60 h
= 1/15 h or (4 $\times $ 60) s
= 240 s
Speed in m/s:
Speed = Distance/Time
= 1500/240
= 6.25 m/s
Speed in km/h:
Speed = Distance/Time
= 1.5/(1/15)
= 22.5 km/h
Page No 185:
Question 16:
A cheetah runs with a speed of 96 km/h. how long would it take to cover 288000 m.
Answer:
Speed of the cheetah = 96 km/h
Distance covered by the cheetah = 288000 m
= (288000/1000) km
= 288 km (∵ 1 km = 1000 m)
We know:
Speed = Distance/Time
∴ Time = Distance/Speed
= 288 / 96
= 3 h
Page No 185:
Question 17:
What is the speed of a swimmer if she covers 100 m in 60 seconds?
Answer:
Distance covered by the swimmer = 100 m
Time taken by the swimmer = 60 s
∴ Speed = Distance/Time
= 100/60
= 1.67 m/s
Page No 185:
Question 18:
How far would a car travel in 4 seconds at a speed of 45 km/h?
Answer:
Given:
Speed of the car = 45 km/h
$=\frac{45\times 1000}{1\times 3600}$
= 12.5 m/s (∵ 1 km = 1000 m and 1 h = 3600 s)
Time taken by the car = 4 s
Using the formula, we get:
Speed = Distance/Time
Or,
Distance = Speed $\times $ Time
= 12.5 $\times $ 4
= 50 m
Thus, the car would travel a distance of 50 metres in 4 seconds.
Page No 185:
Question 19:
Prashant cycled 100 m in 20 seconds. At this rate how long will it take him to go 1 km? What is his speed in km/h? What distance will he cover in 40 minutes?
Answer:
Distance travelled by Prashant in 20 s = 100 m
∴ Distance travelled by Prashant in 1 s = 100/20
= 5 m
Speed of Prashant = 5 m/s
Now,
Time taken by Prashant to travel a distance of 5 m = 1 s
∴ Time taken by him to travel a distance of 1 km or 1000 m $=\frac{1}{5}\times 1000$
$=200\mathrm{s}$
Speed in km/h
5 m/s $=\frac{(5/1000)}{1/3600}$
$=18\mathrm{km}/\mathrm{h}$ (∵ 1 m = 1/1000 km and 1 s = 1/3600 h)
Again,
Distance covered in 1 s = 5 m
∴ Distance covered in 40 min or 2400 s = 5 $\times $ 2400
= 12000 m or 12 km
Therefore, Prashant will cover 12 kilometres in 40 minutes.
Page No 186:
Question 20:
Point out the mistakes in the following cases.
(a) Two cars move for 5 minutes and 2 minutes respectively. The second car is faster because it takes less time.
(b) Two cars move at a speed of 45 km/h. Their velocities are the same.
(c) Trains moving in the same direction have the same velocity.
(d) A motorist travels 600 km. Another motorist travels only 560 km. the second car is slower than the first.
Answer:
(a) The distances travelled by the two cars are not given; these are required to calculate the speeds of the cars and to evaluate which car is faster. Hence, we cannot say that the second car is faster than the first car.
(b) If both the cars are moving at a speed of 45 km/h, then it is not necessary that their velocities are also the same because the directions of motion are not given.
(c) For the same velocity, the magnitudes and directions must be the same, but only direction is given in the statement.
(d) Time, which is required to calculate the speeds of the cars and to evaluate which motorist is slower, is not given. Hence, we cannot say that the second car is slower than the first.
Page No 186:
Question 21:
Sarang wanted to study how fast snails can move. To do this he placed four snails next to each other and marked their trails. He put a cross (x) where each snail had reached after 20 seconds.
(a) Which snail went fastest?
(b) If snail C went on at the same speed for another 10 seconds how far would it go beyond point X?
figure
Answer:
Distance travelled by snail A = 70 mm
= (70/1000) m
= 0.07 m
Time taken = 20 s
∴ Speed of snail A = Distance travelled/Time taken
= 0.07/20
= 0.0035 m/s
Similarly,
Speed of snail B = 0.05/20
= 0.0025 m/s
Speed of snail C = 0.06/20
= 0.0030 m/s
Speed of snail D = 0.04/20
= 0.0020 m/s
(a) Snail A moved the fastest.
(b) Speed of snail C = 0.0030 m/s
Time = 10 s
Distance travelled = Speed $\times $ Time
= 0.0030 $\times $ 10
= 0.03 m
Thus, snail C would go a distance of 0.03 m or 30 mm beyond point X.
Page No 186:
Question 22:
Study the table given below and the answer the questions that follow.
City | Distance From Delhi | Time Taken | Aircraft |
Agra | 360 km | _______h | Pushpak |
Nagpur | 765 km | 3 h 45 m | Dakota |
Mumbai | ___________km | 1 h 50 m | Airbus |
Kolkata | 1035 Km | 1 h 55 m | Boeing 737 |
Chennai | 1860 m | __________ | MIG |
(a) Calculate the speeds of a Boeing 737 and a Dakota.
(b) Calculate how long a Pushpak travelling at 200km/h will take to cover the distance between Delhi and Agra?
(c) An Airbus travels at the same speed as s Boeing 737. What is the distance between Delhi and Mumbai?
(d) How long will a MIG with the speed of 1800 km/h take to travel from Delhi to Chennai?
Answer:
(a) For Boeing 737:
Distance travelled = 1035 km
Time = 1 h 55 min
= 1 h + (55/60) h
= 23/12 h
Speed of Boeing 737 = Distance travelled/Time taken
= 1035/(23/12)
= 955.4 km/h
For Dakota:
Distance travelled = 765 km
Time = 3 h 45 min
= 3 h + (45/60) h
=15/4 h
Speed of Dakota = 765/(15/4)
= 204 km/h
(b) Speed of Pushpak = 200 km/h
Distance travelled = 360 km
Time taken by Pushpak = Distance/Speed
= 360/200
= 1.8 h
(c) Speed of the airbus = Speed of Boeing 737 = 955.4 km/h
Time taken = 1 h 50 min
= 11/6 h
∴ Required distance = Speed $\times $ time
= 955.4 $\times $ 11/6
= 1751.6 km
(d) Speed of MIG = 1800 km/h
Distance travelled = 1860 m
= 1.860 km
∴ Time = Distance/Speed
= 1.860/1800
= 0.001 h
Page No 186:
Question 23:
The graph shows the journeys of Raj and Suraj.
graph
(a) What was the speed of Raj in the first five minutes?
(b) Both of them felt hungry and stopped at Ramji's Mithaiwala. For how long did they stop at the shop?
(c) How far from the start did Raj meet Suraj?
(d) How long did they walk together?
Answer:
(a) Distance travelled by Raj = 500 m
Time taken = 5 min
= 5 $\times $ 60 s
= 300 s
∴ Speed = Distance/Time
= 500/300
= 1.67 m/s
(b) Time that they spent at the shop = 35 $-$ 15
= 20 min
They stopped at the shop for 20 minutes.
(c) Thousand metres far from the start
(d) Distance that they walked together = 1500 $-$ 1000
= 500 m
They walked together a distance of 500 metres.
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