the
following graph and
the following
questing
Among
which
Car a and Car b, which one faster
and whose acceleration
is more
Answers
Answer:
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Thursday, 25 August 2011
CBSE Class IX ( 9th) Science | Chapter 8. Motion | Lesson Exercises
Questions with in the Chapter
Page 100 (CBSE Class IX ( 9th) Science Textbook - Chapter 8. Motion )
Question 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer : Yes, an object even after it has moved through a distance, it can have zero displacement. As we know distance is just length of the path an object has covered irrespective of its direction or position with reference to certain point, where as the shortest distance measured from the initial to the final position of an object is known as the displacement.
For example, an object starts from point A and after covering a distance of say 50 meters, reaches at point B. Here after, it again moves back to point A.
Here the distance covered by object is = AB + BA = 50 m + 50 m = 100 m
where as displacement of object is = AB - BA = 50 m - 50 m = 0 m
As initial position of object is same as that of its final position hence its displacement, which is distance measured from the initial to the final position, is zero.
A
>-----------50 m------------->
<-----------50 m-------------< B
Question 2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Answer : Suppose, a farmer moves along the boundary of a square field of side 10 m in 40 s as shown in the figure given below.
Distance cover by the farmer as he moves from A to B to
C to D to A, along the boundary wall of square field = Perimeter of Square field
= 4 x side of square field
= 4 × 10 m
= 40 m
∴ speed of farmer = 40 m/40 s
= 1 m/s
Distance covered by farmer in 2 minutes 20 seconds = Speed × Time
= 1 m/s × [(2×60) s + 20 s]
= 140 m
Number of round in covering 40 m of distance
along the boundary wall = 1 round
∴ Number of round in covering 140 m of distance
along the boundary wall = 1×140 /40 rounds
= 3.5 round
= 3 1/2 rounds
Which means the farmer will be at point C just diagonally opposite of point A
∴ Relative Displacement of farmer from point A at the end of 3 1/2 round will be = length of AC
which can be determined by the mathematical theorem as given below :
AC = √AB2 + √BC2
= √102 + √102
=10 √22
= 10 × 1.414 m
= 14.14m
a took more time but less distance
b took less time but covered more distance
actually accelaration graph is vel. and time
but given one is speed or vel. graph
speed of b is more