Lesson 2. Motion
Lesson 2. Motion
I. Multiple choice questions:
1. Slope of the velocity – time graph – gives
- Speed
- Displacement
- Distance
- Acceleration
Ans : Acceleration
2. Which of the following graph represents uniform motion of a moving particle?
Ans : b
3.A body moving with an initial velocity 5ms-1 and accelerates at 2ms-1. Its velocity after 10s is
- 20ms^{-1}
- 25ms^{-1}
- 5ms^{-1}
- 22.55ms^{-1}
Ans : 25ms^{-1}
4. In a 100m race, the winner takes 10s to reach the finishing point. The average speed of the winner is
- 5ms^{-1}
- 20ms^{-1}
- 40ms^{-1}
- 10ms^{-1}
Ans : 10ms^{-1}
5. The area under velocity – time graph represents
- Velocity of the moving object.
- Displacement covered by the moving object
- Speed of the moving object
- Acceleration of the moving object
Ans : Displacement covered by the moving object
6. A car is being driven at a speed of 20ms-1 when brakes are applied to bring it to rest in 5s. the deceleration produced in this case will be
- +4ms^{-2}
- -4ms^{-2}
- -0.2ms^{-2}
- +0.25ms^{-2}
Ans : +4ms^{-2}
7. Unit of acceleration is
- ms^{-7}
- ms^{-2}
- ms
- ms^{2}
Ans : ms^{-2}
8. Which one of the following is most likely not a case of uniform circular motion?
- Motion of the earth around the sun
- Motion of a toy train on a circular track.
- Motion of a racing car on a circular track.
- Motion of hours hand on the dial of the clock.
Ans : Motion of a racing car on a circular track.
9. The force responsible for drying of clothes in a washing machine is
- Centripetal force
- Centrifugal force
- Gravitational force
- Electro static force
Ans : Centrifugal force
10. The centrifugal force is
- Real force
- The force of reaction of centripetal force
- Virtual force
- Directed towards the centre of the circular path.
Ans : Virtual force
11. When a body starts from rest the acceleration of the body after 2 second is ……………. Of its
displacement.
- Half
- Twice
- Four times
- Once fourth
Ans : Half
12. In a 100m race, the winner takes 10s to reach the finishing point. The average speed of the winner is ………………… ms-1
- 5
- 10
- 20
- 40
Ans : 10
II. Fill in the blanks :
1. Speed is a ………………………………………. quantity whereas velocity is a …………………………….. quantity.
Ans : Scalar, Vector
2. The slope of the distance – time graph at any point gives …………………………….. .
Ans : Speed
3. Consider an object is rest at position x = 20m, Then its displacement – time graph will be straight line to
Ans : Parallel
4. Negative acceleration is called …………………………….. .
Ans : retardation (or) deceleration
5. Area under velocity – time graph shows …………………………….. .
Ans : displacement
III. State whether true or false. If false, correct the statement :
1. The motion of a city bus ina heavy traffic road is an example for uniform motion. ( False )
Ans : The motion of a city bus in a heavy traffic road is an example for non-uniform motion.
2. Acceleration can get negative value also. ( True )
3. Distance covered by a particle never becomes zero but displacement becomes zero. ( True )
4. The velocity — time graph of a particle falling freely under gravity would be a straight
line parallel to the x axis. ( False )
Ans : The velocity – time graph of a particle moving at uniform velocity, would be straight line parallel to the x axis.
5. Ifthe velocity – time graph of a particle is a straight line inclined to X-axis then its displacement – time graph will be a straight line. ( True )
IV. Assertion and reason type questions :
1. Assertion : The accelerated motion of an object may be due to change in magnitude of velocity or direction or both of them..
Reason : Acceleration can be produced only by change in magnitude of the
velocity. It does not depend the direction.
- If both assertion and reason are true and reason is the correct explanation of assertion.
- If both assertion and reason are true but reason is not the correct explanation of assertion.
- If assertion is true but reason is false.
- If assertion is false but reason is true
Ans : If assertion is true but reason is false.
2. Assertion : The Speedometer of acar ora motor-cycle measures its average speed.
Reason ; Average velocity is equal to total displacement divided by total time taken.
- If both assertion and reason are true and reason is the correct explanation of assertion.
- If both assertion and reason are true but reason is not the correct explanation of assertion.
- If assertion is true but reason is false.
- If assertion is false but reason is true
Ans : Assertion is false but reason is true
3. Assertion : Displacement of a body may be zero when distance travelled by it is not zero.
Reason : The displacement is the shortest distance between initial and final position.
- If both assertion and reason are true and reason is the correct explanation of assertion.
- If both assertion and reason are true but reason is not the correct explanation of assertion.
- If assertion is true but reason is false.
- If assertion is false but reason is true
Ans : Both assertion and reason are true and reason is the correct explanation of assertion
V. Match the following
1. Motion of a body covering equal distances in equal interval | |
2. Motion with non uniform Acceleration | |
3. Constant retardation | |
4. Uniform acceleration. | |
Ans : 1 – D, 2 – C, 3 – A, 4 – B |
VI. Short questions & answers:
1. Define velocity?
Velocity is the rate of change of displacement. It is the displacement in unit time.
2. Distinguish distance and displacement?
Distance | Displacement |
1. The actual length of the path travelled by a moving body irrespective of the direction. | The change in position of a moving body in a particular direction. |
2. It is a scalar quantity. | It is vector quantity. |
3. What do you mean by uniform motion?
An object is said to be in uniform motion if it covers equal distance in equal intervals of time how so ever bit or small these time intervals may be. A particle is in uniform motion when it moves with constant velocity.
4. Compare speed and velocity?
Speed | Velocity |
1. The rate of change of distance (or) the distance travelled in unit time. | The rate of change of displacement in unit time. |
It is a scalar quantity | It is vector quantity. |
5. What do you understand about negative acceleration?
- If final velocity is less than initial velocity, the velocity decreases with time and the value of acceleration is negative. It is called negative acceleration.
- Negative acceleration is called retardation (or) deceleration.
6. What remains constant in uniform circular motion? And what changes is continuously in
uniform circular motion?
- Object is moving with a constant speed along a circular path.
- The direction changes continuously in uniform circular motion.
7. Is the uniform circular motion accelerated? Give reasons for your answer?
Yes, the uniform circular motion is accelerated.
A body is said to be accelerated if the velocity of the body changes either in magnitude (or) in direction. But in the case, the change in velocity is only due to the change in direction.
8. What is meant by uniform circular motion? Give two examples of uniform circular motion.
When an object moves with constant speed along a circular path, the motion is called uniform circular motion.
Examples:
- Revolution of earth around the sun.
- Revolution of moon around the earth.
- In an atom, an electron moves around the nucleus in a circular path.
VIII. Paragraph Questions
1. Derive equations of motion by graphical method.
Equations of motion from velocity – time graph:
Graph shows the change in velocity with time for an uniformly accelerated object. The object starts from the point D in the graph with velocity u. Its velocity keeps increasing and after time t it reaches the point B on the graph.
The initial velocity of the object = u =OD = EA
The final velocity of the object = v = OC = EB
Time = t = OE = DA
Also from the graph we know that, AB = DC
First equation of motion
By definition, acceleration = change in velocity / time
= (final velocity – initial velocity)/time | |
= (OC – OD) / OE | |
= DC / OE | |
a | = DC / t |
DC | = AB = at |
From the graph EB | = EA + AB |
v = u + at ………………………………….. (1) |
This is first equation of motion.
Second equation of motion
From the graph the distance covered by the object during time t is given by the area of quadrangle DOEB
s | = area of the quadrangle DOEB |
= area of the rectangle DOEA + area of the triangle DAB | |
= (AE × OE) + (1/2 × AB × DA) | |
s = ut + ½ at^{2} ………………………………… (2) |
This is second equation of motion.
Third equation of motion
From the graph the distance covered by the object during time t is given by the area of the quadrangle DOEB. Here DOEB is a trapezium. Then
S | = area of trapezium DOEB |
= ½ × sum of length of parallel side × distance between parallel sides | |
= ½ × (OD + BE) × OE | |
S | = ½ × (u + v) × t |
since a | = (v – u) / t or t = (v – u)/a |
Therefore s | = ½ × (v + u) × (v – u)/a |
2as | = v ^{2} – u^{2} |
v2 = u2 + 2 as …………………………………. (3) |
This is third equation of motion.
IX. Exercise Problems
2. A ball is gently dropped from a height of 20m. If its velocity increases uniformly at the rate of 10 ms^{-2} with what velocity will it strike the ground? After what time will it strike the ground?
Here we have
Initial velocity | = 0 |
Distance, s | = 20m |
Acceleration, a | = 10m/s^{2} |
Final velocity, v | = ? |
Time, t | = ? |
a) Calculation of final velocity,
We know that,v^{2} | = u^{2}+2as |
v^{2} | = 0+2~x 10 m/s? x 20m |
v^{2} | = 400 m^{2}/s^{2} |
= √400m^{2} /s^{2} | |
v | = 20 m/s |
b) Calculation of time, t
We know that,v | = ut+at |
20m/s | = 0+10m/s^{2} x t |
t | 20m/s^{2 }/ 20m/s = 2s |
‘. Ball will strike the ground at a velocity of 20 ms^{-1}
Time taken to reach the ground = 2s. |
2. 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 m and 20 s?
Ans. Here we have
Diameter | = 200m |
Radius | = 200 m/2 = 100 m |
Time of one rotation | = 40s |
Time after 2m 20 s | 2 x 60 s +20 s= 140 s |
Distance after 140 s | = ? |
Displacement after 140 s | = ? |
Circular track with diameter of 200m |
We know that, velocity
along a circular path | = circumference / time |
v | = 2mr / 40 s |
v |
= 2 x 3.14 x 100 m / 40 s |
v | = 628 m / 40 s |
v | = 15.7 m/s |
a) Distance after 140 s
We know that, distance | = velocity x time |
=> Distance | = 15.7 m/s x 140s |
= 2198m |
b) Displacement after 2 min 20s i.e, in 140s
We know that, distance | = velocity x time |
Since, rotation in 4o s | = 1 |
ஃ Rotation in l s | = 1 / 40 |
ஃ Rotationin 140 s | = 1 / 40 x 140 = 3.5 |
ஃ In 3.5 rotation athlete will be just at the opposite side of the circular track.
i.e. at a distance equal to the diameter of the circular track which is equal to 200 m
ஃ Distance covered in 2 min 20 s = 2198 m
Displacement after 2 min 20 s = 200 m.
3. Aracing car has a uniform acceleration of 4ms?. What distance it covers in 10s after the start?
Here we have
Acceleration, a | = 4 m/s^{2} |
Initial velocity u | = 0 |
Time t | = 10 s |
Distance (s) covered | = ? |
We know that, s | = uf +½at^{2} |
s | = (0 x10s) + [½ x 4 m/s? x (10s)^{2}] |
= ½ x 4 m/s^{2} x 100s^{2} | |
= 2x 100 m=200m | |
Thus, racing car will cover a distance of 200 m after start in 10s with given acceleration. |
பயனுள்ள பக்கங்கள்