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Question 1
Below Figure represents the speed time graph for a particle. Find the distance covered by the particle
a. between t= 30min. and t= 40min.
b. between t=10 min and t=40 min
c. what is the speed of the particle
Answer
a. Time taken =10 min = 10/60 hr
Speed of the graph =15 km/hr
Distance =Speed X time taken =15 x 10/60= 2.5 Km
b. Time taken =30 min = 30/60 hr
Speed of the graph =15 km/hr
Distance =Speed X time taken =15 x 30/60= 7.5 Km
c) 15 km/hr
Very Important question
Question 2
A particle starts from rest and moves with a uniform acceleration of 5m/s
2 for 10s and then it moves with a constant velocity for 4s. Later it slows down and comes to rest in 5s. Draw the velocity graph for the motion of the body and answer the following questions:
a. What is the maximum velocity attended by the body?
b. What is the distance travelled during this period of acceleration?
c. What is distance travelled when the body was moving with constant velocity?
d. What is the retardation of the body while slowing down?
e. What is the distance travelled by retarding?
f. What is the total distance travelled?
Answer
Draw the velocity time graph as shown below
(a) Maximum velocity will be reached when acceleration is stopped at the end of first 10 sec and it is shown clearly in graph
v=50 m/s
(b) Distance is given by the area of graph enclosed till 10 sec
s= (1/2)X 50 X10 = 250 m
(c) distance travelled when the body was moving with constant velocity is given by the enclosed graph from A to B
= 4X50 =200m
(d)Retardation is given by the slope of the curve from B to C
a=(0-50)/5= -10m/s
2
(e) Distance = (1/2)X 50 X5 =125 m
(f) Total distance is given by the total area of the graph = 250+200+125=575m
Question 3
Two object are thrown vertically upwards simultaneously with their velocities x and y respectively. Prove that
a. The heights reached by them would be in the ratio of x
2: y
2
b. The time taken to reach the maximum height would be in the ratio of x:y
(Assume upward acceleration is –g and downward acceleration to be +g).
Answer
From Third equation of motion
v2=u2+2as
Here u=x and a=-g,v=0
0=x2-2gs
s1=x2/2g
Similarly
for u=y and a=-g,v=0
0=y2-2gs
s2=y2/2g
So ratio of Height
s1/s2 = x2: y2
Now from Ist equation of motion
v=u+at
0=x-gt1
or t1=x/g
Similarly
t2=y/g
t1/t2 = x: y
Question 4
True and false questions
a. Displacement cannot be zero
b. Average speed= Total distance/time
c. Average velocity = Total displacement /time
d. slope of distance-time graph indicates the speed
e. It is possible to have Object moving with uniform speed but variable acceleration.
f. It is possible to have Object moving with uniform velocity but non-uniform acceleration.
Answer
a.false
b. True
c. True
d. True
e. True
f. False
Question 5
Find the distance covered by a particle during the time interval t=0 and t=4s for which the speed time graph is shown in figure
Answer
Distance is given by the area enclosed by speed-time graph
Distance = Area = (1/2) X20 X4 = 40 m
Question 6
Figure shows distance time graph of three objects A ,B and C
a. which object is moving with a greater speed
b. which object is moving with slowest speed
Answer
a. Greatest speed is the greatest slope .So A is the answer
b. Lowest speed is the lowest slope .So C is the answer
Question 7
Marc runs from one end to the other end of a semicircular track whose radius is 140m. What is the distance covered by the marc and what is his displacement?
Answer
Distance = 3.14 X 140 = 439.6 m
Displacement =2r =280m
Question 8
A jogger moves 500m in 2 minutes and next 1000m in 30s on the same straight path. What is his average speed and average velocity?
Answer
Total distance = 500 + 1000=1500 m
Total Time= 120 + 30 =150 s
Average Speed = Total distance /Total Time = 10 m/s
Average velocity is same as Average speed in this case as motion is in straight line
Question 9
A big truck moving along a straight line at a speed of 54km/hr stop in 5s after the breaks are applied.
- Find the acceleration, assuming it to constant.
- Plot the graph of speed versus time.
- Using the graph. Find the distance covered by the car after the brakes are applied?
Answer
a. u=54 km/hr=15 m/s , t= 5 s ,v=0 ,a=?
$v=u+at$
a= -3 m/s
2 ( Negative as retardation)
b. Speed -time graph is shown below
c. Now distance covered is equal to area enclosed by the speed time graph
Area =$\frac {1}{2} \times 15 \times 5 = 37.5 m$
A swimmer swims 90m long pool. He covers the distance of 180m by swimming from one end to other end back along the same path. If he covers the first 90m at speed of 2m/s, then how fast he swim so that his average speed is 3m/s?
Answer
6 m/s
Question 11
a. Write the difference between Distance and displacement
b. Write the difference between Uniform and Non uniform Motion
Question 12
Lori moves 4m due east and then 3m due west.
- What is the distance covered by the Lori?
- What is its displacement?
Answer
Distance = 3+4 = 7 km
Now Displacement = 4 -3 =1 km as 3 km is in backward direction
Question 13
A car moves at a speed of 40km/h, It is stopped by applying brakes which produces a uniform acceleration of -0.6m/s
2. How much distance will the vehicle move before coming to stop?
Answer
u=40 km/h=11.11 m/s ,a=-0.6m/s2,v=0
$v^2=u^2 + 2as$
s=102.87 m
Question 14
A person walks along the sides of a square field. Each side is 10m long. What is the maximum magnitude of displacement of the person in any time interval?
Answer
Maximum Displacement happened when it is diagonally across
$d= a \sqrt {2}= 10 \sqrt {2}$
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