CHAPTER - 3 MOTION
Date : 21.06.16
Question: Define the scalar quantity and name 5 physical quantities which are scalar .
Answer : Scalar quantity is the physical quantity in which only magnitude is measured and direction is not specified . For eg : speed , distance,time.
These all quantities are scaler because there direction is not specified only their magnitude is measured .
Date : 24 .06.16
Graph :
Graph of the above table
Question: Define the scalar quantity and name 5 physical quantities which are scalar .
Answer : Scalar quantity is the physical quantity in which only magnitude is measured and direction is not specified . For eg : speed , distance,time.
These all quantities are scaler because there direction is not specified only their magnitude is measured .
Date : 24 .06.16
Graph :
Graph of the above table
It is not a uniform motion
Examples
Example 1 :
Question : An object travels 16 m in 4 s and then another 16 m in 2 s . What is the average speed of the object ?
Answers : The total distance covered by the object = 16 + 16 = 32 m
Total time taken by the object = 4 s + 2 s = 6 s
average speed = Total distance travelled = 32 m = 5.33 m/s
Total time taken 6 s
So , the average speed of the object is 5.33 m/s
Question : Usha swims in 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 .
Answer : The total distance covered by Usha in 1 minute = 90 + 90 = 180
displacement of Usha in 1 min = 0 m
Average speed = Total distance covered
Total time taken
= 180 m = 180 m x 1 min
1 min 1 min 60 secs
= 3 m/s
Average velocity = Displacement
Total time taken
= 0 m
60 s = 0 m/s
so , we find that the average speed of Usha is 3 m/s and her average
velocity is 0 m/s .
THEORETICAL 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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