CHAPTER-3
MOTION
Questions:-
21-JUNE-2016
1) Define the scalar quantity and
name 5 physical quantities which are scalar and give reasons why these physical
quantities are scalar?
ANS- Scalar quantity is the physical quantity in which only
magnitude is measured and
direction is not specified .
Eg- Time, Speed, Volume,
Temperature, Area.
All these quantities
are scalar quantities because there direction is not specified, their only
magnitude is measured
24-JUNE-2016
2) GRAPH
Graph of the above table:-
It is a non-uniform motion.
EXAMPLES
(Both
theoretical and numerical questions)
1. An object travels
16 m in 4 s and then another 16 m in 2 s. What is the average speed of the
object?
ANS- Total distance covered by the object = 16+16 = 32m
Total time taken = 4+2
= 6s
Therefore, average
speed of the object = 32 =
5.33 m s-1
6
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
trip took 8 h, calculate the average speed of the car in km
h–1 and m s–1.
ANS- Total distance covered by
the object = 2400-2000 = 400km
Total time taken = 8h
Therefore, average
speed of the object = 400 = 50 km h-1
8
In m s–1 = 13.9 m s–1.
3 . Usha swims in a 90 m long
pool. She covers 180 m in one minute by swimming from one end to
the other and back along the same straight path. Find the average
speed and average velocity of Usha.
ANS- Total distance covered by Usha in 1 min is 180 m.
Displacement of Usha in
1 min = 0 m
Average speed of the
object = 180 = 3 m s-1
60
Average velocity = Displacement = 0m = 0 m s-1
Total Time 60
4.Starting
from a stationary position, Rahul paddles his bicycle to attain a velocity of 6
m s–1 in 30 s. Then he applies brakes such that the velocity of
the bicycle comes down to 4 m s-1 in the next 5 s.
Calculate the acceleration of the bicycle in both the cases.
ANS- In the first case
U=0 , V=6 m s-1 , t = 30sec ,
a = ?
a = V-U = 6-0 = 1 m
s-2 = 0.2 m s–2
t
30 5
In the second case
U=6 m s-1 , V=4 m s-1 , t =
5sec , a = ?
a = V-U
= 4-6 = -2 m s-2 = -0.4
m s–2
t
5 5
5. A train starting from rest attains a velocity
of 72 km h–1 in 5 minutes. Assuming that the
acceleration is uniform, find (i) the
acceleration and (ii) the distance travelled by the train for
attaining this velocity.
ANS- U=0, V= 72 km h-1 = 20 m s-1, t = 5min = 300sec
(i) a = V-U = 20-0 =
1 m s-2
t
300 15
(ii) by IInd equation of motion
S= ut + 1/2 at*2
S= 0X300 + 1/2 X 1/15 X
90000
S= 3000m = 3km
6. A car accelerates uniformly from 18 km h–1 to 36 km h–1 in
5 s. Calculate
(i) the acceleration and
(ii) the distance covered by the car in that time.
ANS- U= 18 km h-1 = 5 m s-1 , V= 36 km h-1 = 10 m s-1 , t = 5 s
(i) a = V-U = 10-5 =
1 m s-2
t
5
(ii) by II nd equation of motion
S= ut + 1/2 at*2
S= 5X5 + 1/2 X 1 X 25
S= 25 + 12.5 = 37.5 m
7. The brakes applied to a car produce an
acceleration of 6 m s-2 in the opposite direction to the
motion. If the car takes 2 s to stop after the application
of brakes, calculate the distance it travels during
this time.
ANS- a = -6 m s-2 , t = 2 sec , V= 0 m s-2
by I st equation of
motion
V=U+at
0= U+(-12)
0=U-12
U= 12 m s-1
by II nd equation of motion
S= ut + 1/2 at*2
S= 12X2 + 1/2 X -6 X 4
S= 24 - 12 = 12 m
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