Page 93 - Understanding NCERT Science 09

(ii) From Eq. (7.7) we have Example 7.7 The brakes applied to a car
2 a s = v – u = v – 0 produce an acceleration of 6 m s in the
2
2
2
-2
Thus,
opposite direction to the motion. If the
2 car takes 2 s to stop after the application
v
s = of brakes, calculate the distance it
2a travels during this time.
–1 2
(20 m s ) Solution:
=
–2
2×(1/15) m s We have been given
–1
a = – 6 m s ; t = 2 s and v = 0 m s .
–2
= 3000 m From Eq. (7.5) we know that
= 3 km
v = u + at
1 0 = u + (– 6 m s ) × 2 s
–2
The acceleration of the train is m s – 2 –1
15 or u = 12 m s .
and the distance travelled is 3 km. From Eq. (7.6) we get
1
s = u t + a t 2
Example 7.6 A car accelerates uniformly 2
–1 –1
from 18 km h to 36 km h in 5 s. 1
–1
–2
Calculate (i) the acceleration and (ii) the = (12 m s ) × (2 s) + 2 (–6 m s ) (2 s) 2
distance covered by the car in that time. = 24 m – 12 m
= 12 m
Solution:
Thus, the car will move 12 m before it
We are given that stops after the application of brakes. Can
–1
u = 18 km h = 5 m s –1 you now appreciate why drivers are
v = 36 km h = 10 m s and cautioned to maintain some distance
–1
–1
t = 5 s . between vehicles while travelling on the
road?
(i) From Eq. (7.5) we have
v – u
a = uestions
t
-1
10 m s – 5 m s -1 Q 1. A bus starting from rest moves
= with a uniform acceleration of
5s 0.1 m s for 2 minutes. Find (a)
-2
= 1 m s –2 the speed acquired, (b) the
(ii) From Eq. (7.6) we have
distance travelled.
1 2. A train is travelling at a speed
s = u t + a t 2 –1
2 of 90 km h . Brakes are applied
so as to produce a uniform
1 acceleration of – 0.5 m s . Find
-2
–2
= 5 m s × 5 s + × 1 m s × (5 s) 2
–1
2 how far the train will go before it
= 25 m + 12.5 m is brought to rest.
3. A trolley, while going down an
= 37.5 m
–2 inclined plane, has an
The acceleration of the car is 1 m s and acceleration of 2 cm s . What will
-2
the distance covered is 37.5 m.
be its velocity 3 s after the start?

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