Motion in a Straight Line
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Question : 14
Total: 27
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h– 1 – 1
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min?
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min?
Solution:
Here, distance of the market from the home of the man,
S = 2.5 k m
Speed of the man, while going from his home to the market,
v 1 = 5 k m h − 1
Therefore, time taken to reach the market,
t 1 =
=
= 0.5 h = 30 m i n
Speed of the man, while returning from the market,
v 2 = 7.5 k m h − 1
Therefore, time taken to return the home,
t 2 =
=
=
h = 20 m i n
(i) Over the interval of time 0 to 30 min :
During this interval, the man covers distance from his home upto themarket.
Therefore, displacement = 2.5 km; and distance covered = 2.5 km
Now, time taken = 30 min = 0.5 h
Therefore, magnitude of the average velocity
=
=
= 5 k m h − 1
average speed=
=
= 5 k m h − 1
(ii) Over the interval of time 0 to 50 min :
During this time interval, the man returns his home.
Therefore, displacement = 0
and distance covered = 2.5 + 2.5 = 5 km
Now, time taken= 50 m i n =
=
h
Therefore, magnitude of the average velocity=
= 0
and average speed=
= 6 k m h − 1
(iii) Over the interval of time 0 to 40 min :
During the first 30 min of this interval, the man covers a distance of 2.5 km and in next 10 min, he covers 2.5/2 i.e. 1.25 km of the return journey.
Therefore, displacement = 2.5 – 1.25 = 1.25 km;
and distance covered = 2.5 + 1.25 = 3.75 km
Now, time taken = 40 min=
h
Therefore, magnitude of the average velocity
=
=
= 1.875 k m h − 1
and average speed=
= 5.625 k m h − 1
Speed of the man, while going from his home to the market,
Therefore, time taken to reach the market,
Speed of the man, while returning from the market,
Therefore, time taken to return the home,
(i) Over the interval of time 0 to 30 min :
During this interval, the man covers distance from his home upto themarket.
Therefore, displacement = 2.5 km; and distance covered = 2.5 km
Now, time taken = 30 min = 0.5 h
Therefore, magnitude of the average velocity
average speed
(ii) Over the interval of time 0 to 50 min :
During this time interval, the man returns his home.
Therefore, displacement = 0
and distance covered = 2.5 + 2.5 = 5 km
Now, time taken
Therefore, magnitude of the average velocity
and average speed
(iii) Over the interval of time 0 to 40 min :
During the first 30 min of this interval, the man covers a distance of 2.5 km and in next 10 min, he covers 2.5/2 i.e. 1.25 km of the return journey.
Therefore, displacement = 2.5 – 1.25 = 1.25 km;
and distance covered = 2.5 + 1.25 = 3.75 km
Now, time taken = 40 min
Therefore, magnitude of the average velocity
and average speed
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