Looking for NCERT Revision Notes? Check out here.
Q1: An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Ans: Yes, the object may have zero displacement if the object has stopped the same position from where it has started initially.
Q2: A farmer moves along the boundary of a square field of side 10m in 40s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Ans: Given, the farmer travels along the boundary of a square filed of side 10m in 40s.
So, total distance he travelled in 40s = 10 x 4 = 40m (Perimeter of square)
So, 1s he will travel = 40/40 = 1m
At the end of 2 minute 20s he would travel = (2x60 + 20) x 1
= 140m
Q3: Which of the following is true of displacement?
Ans: Both (a) and (b) statements are false. (a) is false for displacement of an object can be zero if its initial and final position is same. (b) false for displacement of an object can be equal to zero but cannot be greater than it.
Q4: Distinguish between speed and velocity.
Ans:
|
Speed |
Velocity |
|
Speed is the distance travelled per unit time. |
Velocity is the displacement per unit time. |
|
It is a scalar quantity. |
It is a vector quantity. |
|
The SI unit of speed is m/s |
The SI unit of speed is m/s |
Q5: Under what conditions is the magnitude of average velocity of an object equal to its average speed?
Ans: The average velocity of an object will be equal to its average speed when the total distance and total displacement travelled by the object is equal. And it is only possible when the object is travelling in a straight line.
Q6: What does the odometer of an automobile measure?
Ans: The odometer measures the total distance travelled by the automobile till date.
Q7: What does the path of an object look like when it is in uniform motion?
Ans: The path of an object with uniform motion looks like a slant straight line.
Q8: During an experiment, a signal from a spaceship reached the ground station in ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light that is, 3x108 m s-1.
Ans:
Given, time t = 5 min and speed of signal = 3x108
We know that
speed = distance / time
=> 3x108 = s / (5x60) seconds
=> 3x108 x 5x60 = s
=> 90 x 109 = s
=> 9 x 1010 = s
So, the distance between spaceship and the ground is 9 x 1010 metres.
Q9: When will you say a body is in (i) uniform acceleration (ii) non-uniform acceleration?
Ans: (i) When the speed of an object increase with constant rate, then it is said to be uniform acceleration.
(ii) When the speed of an object increase with variable rate, then it is said to be non-uniform acceleration.
Q10: A bus decreases its speed from 80 km/h to 60 km/h in 5s. Find the acceleration of the bus.
Ans: Given, initial velocity u = 80 km/h
= (80 x 1000) / 3600
= 22.22 m/s
And final velocity v = 60 km/h
= (60 x 1000) / 3600
= 16.66 m/s
And time t = 5 seconds
Since,
Acceleration a = change in velocity / time taken
= (v - u) / t
=(16.66 – 22.22) / 5
= -6/5
= -1.2 m s-2
Therefore, acceleration of bus is – 1.2 m s-2
Q11: A train starting from a railway station and moving with uniform acceleration attains a speed 40 km/h in 10 minutes. Find its acceleration.
Ans: Given, initial velocity u = 0 km/h = 0 m/s
Final velocity v = 40km/h
= (40 x 1000)/3600
= 11.11 m/s
Time t = 10 min
= 600 seconds
Now, acceleration a = change in velocity / time taken
= (v-u) / t
= (11.11 – 0)/ 600
= 0.018 m/s2
Q12: What is the nature of distance-time graph for uniform and non-uniform of an object?
Ans: The distance time graph shows straight line for uniform motion and a curve line for non-uniform motion.
Q13: What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Ans: There will be no motion for an object whose distance time graph is a straight line parallel to the time axis.
Q14: What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Ans: The object is travelling with constant speed if its speed-time graph is a straight line parallel to the time axis.
Q15: What is the quantity which is measured by the area occupied below the velocity-time graph?
Ans: Displacement is the quantity that is measured by the area occupied below the velocity time graph.
Q16: A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Ans: Given, initial velocity u = 0 m/s
Uniform acceleration a = 0.1 m/s2
Time t = 2 minutes = 120 seconds
(a) Using first equation of motion v = u + at
v = 0 + (0.1 x 120)
v = 12 m/s
Therefore, speed acquired by bus is 12 m/s.
(b) Now, using third equation of motion 2as = v2 – u2
s = (v2 – u2) / 2a
s = [(12)2 – (0)2] / 2(0.1)
s = (144 - 0) /0.2
s = 144/0.2
s = 1440/2
s = 720 m
So, the distance travelled by the bus is 720 m.
Q17: A train is travelling at a speed of 90 km h–1. Brakes are applied so as to produce a uniform acceleration of – 0.5 m s-2. Find how far the train will go before it is brought to rest.
Ans: Given, initial velocity u = 90 km/h
u = 90 x 1000 / 3600
u = 25 m/s
Final velocity v = 0 km/h = 0 m/s
Acceleration a = -0.5 m/s2
Using third equation of motion 2as = v2 – u2
=> 2(-0.5)x s = (0)2 – (25)2
=> -1.0 x s = 0 – 625
=> s = -625 / -1.0
=> s = 625 m
So, the train will travel 625 metres before it is brought to rest.
Q18: A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2. What will be its velocity 3s after the start?
Ans: Given, initial velocity u = 0 cm/s
Acceleration a = 2 cm/s2
Time t = 3s
Using second equation of motion v = u + at
=> v = 0 + (2)(3)
=> v = 6 cm/s
So, the velocity of trolley after 3s will be 6 cm/s.
Q19: A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10s after start?
Ans: Given, acceleration a = 4 m s-2
Initial velocity u = 0 m/s
Time t = 10s
Using second equation of motion s = ut + ½ at2
=> s = (0) (10) + ½ (4)(10) 2
=> s = 0 + ½ 400
=> s = 200 m
So, the racing car will travel distance of 200 m.
Q20: A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Ans: Given, initial velocity u = 5 m/s
Acceleration a = -10 m/s2
Final velocity v = 0 m/s
Using third equation of motion 2as = v2 – u2
=> 2(-10) x s = (0)2 – (5)2
=> -20 x s =0 – 25
=> s = -25/-20
=> s = 1.25 m
Therefore, the height attained by the stone is 1.25m.
Now, using first equation of motion v = u + at
=> 0 = 5 + (-10) x t
=> 0 = 5 -10t
=> -5 = -10t
=> -5/-10 = t
=> 0.5 = t
Therefore, the stone took 0.5 seconds to reach its maximum height.
Q1: An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Ans: Distance:
Given, the diameter of circular track is 200m.
So, radius of circular track = 200/2 = 100m
The time taken by the athlete to complete one round is 40 seconds.
So, number of round he will complete in 140s = 140/40 = 3.5 rounds
Now, total distance travelled by the athlete at the end of 140s
Distance = 2 x 3.14 x 100 x 3.5
Distance = 2200m
So, total distance travelled by the athlete is 2200 m.
Displacement:
If athlete starts from point A, then after 140s he will be at point B.
So, displacement = diameter of the circle
= 200m
Hence, the displacement was 200m.
Q2: Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Ans: Given, the distance between A to B = 300m
And time taken = 2 min 30s = 150s
(a) So, average speed from A to B = total distance travelled / total time taken
= 300/150
= 2 m/s
Now, velocity from A to B = displacement / time taken
= 300/ 150
= 2 m/s
(b) Total distance travelled from A to C = 300 + 100 = 400m
And time taken from A to C = 2min 30s + 1min
= 150s + 60s
= 210s
Now, average speed from A to C = total distance travelled / total time taken
= 400 /210
= 1.904 m/s
Now, displacement from A to C = 300 – 100 = 200m
And velocity from A to C = displacement / time taken
= 200 / 210
= 0.95 m/s
Q3: Abdul, while driving to school, computes the average speed for his trip to be 20 km h–1. On his return trip along the same route, there is less traffic and the average speed is 30 km h–1. What is the average speed for Abdul’s trip?
Ans: Given, the average speed while going to school = 20km/h
Let the distance be d.
So, total distance travelled = d + d = 2d
Average speed = total distance travelled / total time taken
=> 20 = d / time
=> time = d/20
And given the average speed on return trip was 30 km/h
Again, average speed = total distance travelled / total time taken
=> 30 = d / time
=> time = d/30
Now, total time taken by Abdul = d/20 + d/30
= 2d/60 + 2d/30 [Since, LCM of 20 and 30 is 60]
= 5d/60
= d/12
Now, the average speed of Abdul’s trip = total distance travelled / total time
= 2d / (d/12)
= (2d x 12) / d
= 2 x 12
= 24
So, the average speed of Abdul’s total trip is 24 km/h.
Q4: A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s–2 for 8.0 s. How far does the boat travel during this time?
Ans: Given, initial velocity u = 0 m/s
Acceleration a = 3 m/s-2
Time t = 8 seconds
Using second equation of motion s = ut + ½ at2
=> s = (0)(8) + ½ (3)(8)2
=> s = 0 + ½(192)
=> s = 96
So, the distance travelled by motorboat is 96 m
Q5: A driver of a car travelling at 52 km h–1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h–1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Ans: First car:
Given, initial velocity u = 52 km/h
= (52 x1000) / 3600
= 14.4 m/s
Final velocity v = 0 m/s
Time t = 5 seconds
Acceleration a = (v -u) / t
a = (0 -14.4)/5
a = -14.4/5
a = -2.88 m/s2
Using third equation of motion s = ut + ½ at2
=> s = (14.4)(5) + ½ (-2.88)(5)2
=> s = 72 + ½ (-72)
=> s = 72 – 36
=> s = 36
So, the distance travelled by the first car is before stopping is 36 m.
Second car:
Given, initial velocity u = 3 km/h
= (3 x 1000)/ 3600
= 0.83 m/s
Final velocity v = 0 m/s
Time t = 10s
Acceleration a = (v-u) /t
= (0 – 0.83)/10
= -0.083 m/s2
Using third equation of motion s = ut + ½ at2
=> s = (0.83)(10) + ½ (-0.083)(10)2
=>s = 8.3 + ½ (-0.083)(100)
=>s = 8.3 + ½ (-8.3)
=> s = 8.3 - 4.15
=> s = 4.15
So, the distance travelled by the second car before stopping is 4.15 m.
Q6: Fig 8.11 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:
(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?
Ans: (a) A travelled distance from 6 to 12 km in 2 hours
So, speed of A = distance/time
= 6 /2
= 3 km/h
B travelled distance from 0 to 12 in 1.4 hours
So, speed of B = distance/time
= 12/1.4
=8.57 km/h
C travelled distance from 2 to 12 km in 1.6 hours
So, speed of C = distance/ time
= 10/1.6
= 6.25 km/h
Hence, from the above we can say that B is travelling faster.
(b) No, none of the object meet at each other at any point.
(c) The C has travelled 8 km when B passes A.
(d) The B has travelled 5 km when it passes C.
Q7: A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground?
Ans: Given, initial velocity u = 0 m/s
Distance travelled (height) = 20m
Acceleration a = 10 m s-2
Using third equation of motion v2 – u2 = 2as
=> v2 = 2as + u2
=> v2 = 2(10)(20) + (0)2
=> v2 = 400
=> v = 20
So, the velocity of ball when it strike the ground was 20 m/s
Now, using first equation of motion v = u + at
=> 20 = 0 + (10)x t
=> 20/10 = t
=> 2 = t
So, the ball took 2 seconds to strike the ground.
Q8: The speed-time graph for a car is shown is Fig. 8.12.
(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
Ans:
(a) According to the above figure distance travelled car = area of triangle
= ½ x base x height
Here, height = 6 m
And base = 4
So, ½ x 4 x 6
= 4 x 3
= 12
So, the distance travelled by the car is 12 m.
(b) The above figure shows the uniform motion of the car.
Q9: State which of the following situations are possible and give an example for each of these:
(a) An object with a constant acceleration but with zero velocity
(b) An object moving with an acceleration but with uniform speed.
(c) An object moving in a certain direction with an acceleration in the perpendicular direction.
Ans: (a) This situation is possible if a ball is thrown in the sky, after reaching its maximum height its velocity will become zero however, its acceleration (gravitational force) will remain constant.
(b) This situation is impossible because acceleration either increases or decreases the speed of an object but can’t keep the speed uniform.
(c) This situation is possible if an object is in circular motion. In that case the acceleration will be in the perpendicular direction if object is moving in certain direction.
Q10: An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Ans: Given, radius of orbit = 42250km
And time t = 24 hours
So, distance travelled by satellite = circumference of circle
= 2 x Pie x radius
= 2 x 3.14 x 42250
= 2,65,330 km
Now,
Speed = distance / time
= 265330/24
= 11,055.41 km/h
So, the speed of the satellite is 11,055.41 km per hour.
Speed = distance / time
Average speed = total distance travelled / total time taken
Velocity = displacement / time
Average velocity =( initial velocity + final velocity) / 2
Acceleration =(initial velocity – final velocity) / time taken
First equation of motion v = u + at
Second equation of motion s = ut + ½ at2
Third equation of motion 2as = v2 – u2