Page 87 - Understanding NCERT Science 09

Activity ______________ 7.7
km 1000m 1h
= 50 × ×
• At a time when it is cloudy, there may h 1km 3600s
be frequent thunder and lightning. The –1
sound of thunder takes some time to = 13.9 m s
reach you after you see the lightning. The average speed of the car is
–1
–1
• Can you answer why this happens? 50 km h or 13.9 m s .
• Measure this time interval using a
digital wrist watch or a stop watch.
• Calculate the distance of the nearest Example 7.3 Usha swims in a 90 m long
point of lightning. (Speed of sound in pool. She covers 180 m in one minute
air = 346 m s .) by swimming from one end to the other
-1
and back along the same straight path.
uestions
Q 1. Distinguish between speed and Solution:
Find the average speed and average
velocity of Usha.
velocity.
Total distance covered by Usha in 1 min
2. Under what condition(s) is the
is 180 m.
magnitude of average velocity of
an object equal to its average
Total distance covered
speed? Displacement of Usha in 1 min = 0 m
Average speed =
3. What does the odometer of an Totaltimetaken
automobile measure?
180m 180 m 1 min
4. What does the path of an object = ×
=
look like when it is in uniform 1min 1min 60s
motion? = 3 m s -1
5. During an experiment, a signal
from a spaceship reached the Displacement
ground station in five minutes. Average velocity = Totaltimetaken
What was the distance of the
spaceship from the ground 0m
station? The signal travels at the =
60 s
speed of light, that is, 3 × 10 8
–1
m s . = 0 m s –1
The average speed of Usha is 3 m s –1
–1
and her average velocity is 0 m s .
Example 7.2 The odometer of a car reads
2000 km at the start of a trip and
2400 km at the end of the trip. If the 7.3 Rate of Change of Velocity
trip took 8 h, calculate the average
–1
–1
speed of the car in km h and m s . During uniform motion of an object along a
straight line, the velocity remains constant
Solution:
with time. In this case, the change in velocity
Distance covered by the car, of the object for any time interval is zero.
s = 2400 km – 2000 km = 400 km However, in non-uniform motion, velocity
Time elapsed, t = 8 h varies with time. It has different values at
Average speed of the car is, different instants and at different points of
the path. Thus, the change in velocity of the
s 400 km
v = = object during any time interval is not zero.
av
t 8 h Can we now express the change in velocity of
= 50 km h –1 an object?

76 SCIENCE

Rationalised 2023-24