I. Very Short answer type questions:
1. What are the uses of the distance-time graph?
Ans. The nature of the motion of an object can be studied by plotting of distance-time graph.
Ans. The nature of the motion of an object can be studied by plotting of distance-time graph.
2. Give an example in which distance is equal to the magnitude of displacement.
Ans. when an object is thrown upwards, the distance and displacement will be equal till the time of ascent and when maximum height is attained by the object.
3. The orbit of a satellite at distance 42260 km from the earth is circular. It completes one revolution around the earth in 24 hr. Calculate its linear speed.
Ans. Linear speed = Circumference/time = 2πr/t [ Where r = radius]
= (2 x 3.14 x 42260) / 24
= 11058.03 km/h.
3. The orbit of a satellite at distance 42260 km from the earth is circular. It completes one revolution around the earth in 24 hr. Calculate its linear speed.
Ans. Linear speed = Circumference/time = 2πr/t [ Where r = radius]
= (2 x 3.14 x 42260) / 24
= 11058.03 km/h.
4. Is speed a scalar or a vector quantity?
Ans. Yes, speed is a scalar quantity. It has only magnitude.
5. Is displacement a vector quantity or scalar quantity?
Ans Displacement is a vector quantity.
6. If the displacement of a body is zero is it necessary that the distance covered by it is also zero?
Ans. No, it is not necessary because the distance covered by body is path length and it is always positive.
7. When do the distance and displacement of a moving object have the same magnitude?
Ans. If the motion is along a fixed direction, the distance and displacement of a moving object have the same magnitude.
8. What are the characteristics of the distance-time graph for an object moving at a uniform speed?
Ans. Straight-line inclined to the time axis.
9. What do you mean by displacement?
Ans. The shortest distance traveled by a body in the direction from the initial position to the final position.
10. Is displacement a scalar quantity?
Ans. No, displacement is a vector quantity as it has both direction and magnitude.
11. When do the displacement and distance of a moving body have the same value?
Ans. When the object moves along the straight line in the same fixed direction.
12. What is the S.I unit of speed?
Ans. S.I unit of speed is m/s.
13. What does the speedometer measure?
Ans. Speedometer measures instantaneous speed.
14. What does the area under the velocity-time graph give?
Ans. It gives the magnitude of displacement of the object.
15. A bus takes 8 hours to cover a distance of 320 kilometers. What is the average speed of the bus?
Ans. Average speed = Total distance/ Total time = 320 / 8 = 40 km/h.
16. What is the slope of the body when it moves with uniform velocity?
Ans. The slope is the acceleration.
17. What does the slope position-time graph give?
9. What do you mean by displacement?
Ans. The shortest distance traveled by a body in the direction from the initial position to the final position.
10. Is displacement a scalar quantity?
Ans. No, displacement is a vector quantity as it has both direction and magnitude.
11. When do the displacement and distance of a moving body have the same value?
Ans. When the object moves along the straight line in the same fixed direction.
12. What is the S.I unit of speed?
Ans. S.I unit of speed is m/s.
13. What does the speedometer measure?
Ans. Speedometer measures instantaneous speed.
14. What does the area under the velocity-time graph give?
Ans. It gives the magnitude of displacement of the object.
15. A bus takes 8 hours to cover a distance of 320 kilometers. What is the average speed of the bus?
Ans. Average speed = Total distance/ Total time = 320 / 8 = 40 km/h.
16. What is the slope of the body when it moves with uniform velocity?
Ans. The slope is the acceleration.
17. What does the slope position-time graph give?
Ans. The slope of the position-time graph gives the magnitude of motion.
18. What does the slope velocity-time graph give?
Ans. The slope of the velocity-time graph gives the displacement.
19. When a moving body comes to rest, then what its acceleration is: positive/ negative / zero.
Ans. Negative.
20. What does the odometer of an automobile measure?
Ans. Odometer measures distance traveled by automobile.
20. What does the odometer of an automobile measure?
Ans. Odometer measures distance traveled by automobile.
II.Short Answer Type Questions:
1. Are rest and motion relative terms?
Ans. Yes, rest and motion are relative terms.
While sitting on a moving bus our distance from the walls, roof, and floor of the bus does not change. So with respect to the bus our position does not change. Therefore we are at rest with respect to the bus but our distance from the bus stand changes with time. So we are moving with respect to the bus stand. So motion and rest are relative terms.
Ans. Yes, rest and motion are relative terms.
While sitting on a moving bus our distance from the walls, roof, and floor of the bus does not change. So with respect to the bus our position does not change. Therefore we are at rest with respect to the bus but our distance from the bus stand changes with time. So we are moving with respect to the bus stand. So motion and rest are relative terms.
2. Plot the velocity-time graph showing +a and –a. How is the distance calculated from the velocity-time graph?
Ans.
By finding the area under the velocity-time graph which is equal to displacement we can calculate distance.
Ans. Differences between speed and velocity:
(i) Speed is the rate of change of motion and velocity is the rate of change of motion in a particular direction.
(ii)Speed is scalar quantity and velocity is a vector quantity.
(ii)Speed cannot be zero or negative but velocity can be zero, positive or negative.
4. Why circular motion is accelerated motion? Write an expression of circular motion.
Ans. A particle moving in a circular path changes its direction continuously. its velocity therefore not constant even if its speed is constant. so it is an example of accelerated motion.
Expression of circular motion (v) = 2πr/t Where r= radius and t = time.
5. What is meant by uniform circular motion? Give its two examples.
Ans. If a body moves in a circular path and covers equal distance in equal interval of time, motion is called uniform circular motion.
Two examples are: Revolution of earth and the revolution of the moon.
6. Calculate the acceleration of a body which starts from rest and travels 87.5 min 5 sec.
Ans. The particle starts from rest, so u= 0
Distance traveled by body = 87.5m
Time taken = 5 sec.
S = ut + 1/2 at2
87.5 = 1/2 x a x 5 x 5
a = 87.5 x 2
5 x 5
= 175/25
= 7 m/s2.
7. A car traveling at 36 km/hr speed up to 72 km/hr in 5 sec. What is the acceleration? If the same car stops in 20 sec, what is the retardation?
Ans. Here Initial velocity u = 36 Km/h = 10 m/s and v = 72 Km/h= 20 m/s
Time t = 5 sec.
Acceleration a = (v - u) / t = (20 - 10)/5 = 10 / 5 = 2 m/s2.
If the same car stops in 20 sec then retardation = (v-u) / t
(v=0 m/s and u = 10 m/s)
Retardation = 0 - 10/20 = -1/2 m/s^2= - 0.5 m/s2.
8. A car moves 100 m due east and then 25 meters due west.
(i) What is the distance covered by the car?
(ii) What is its displacement?
Ans. (i) Distance covered by car = 100 + 25 = 125m
(ii) Displacement = 100 - 25 = 75m.
9. A person walks along the side of a square field. Each side is 100m long. What is the maximum magnitude of displacement of the person if any time interval?
Ans. In a square field diagonal is the longest length, so diagonal is the maximum displacement.
If side = 100 m diagonal of sq. = √2 a = √2 x 100 = 141.42 m.
11. The maximum speed of a train is 80 km/h. It takes 10 hours to cover a distance of 400 km. Find the ratio of its maximum speed to its average speed.
Ans. Maximum speed = 80 km/h.
Average speed = Total distance / Total time = 400/10 = 40 m/s
Now the ratio of its maximum speed to its average speed
= 80/40 = 2:1.
12. Given below are few speed time graphs for the motion of objects moving along a straight line which of these graphs represent the motion of the body whose speed is :(i) Increasing with time,(ii) Uniform and (iii) nonuniform.
A B C
Ans. (i) Increasing with time - Graph C
(ii) Uniform - Graph A
(iii) Nonuniform - B
13. Define scalar and vector quantity?
Ans. Scalar is a quantity that has only magnitude or numerical value.
Ex- Temperature and mass are scalar quantity.
A vector is a quantity that has both magnitude and direction.
Ex-Velocity and acceleration is a vector quantity.
14. A body is moving along a circular path of radius R. What are the values of distance and displacement in the half revolution
Ans. If the radius of the circular path is R then circumference = 2πR
For half revolution:
Distance = 1/2 (2πR) = πR
Displacement = diameter = 2R.
15. What is the displacement of a satellite when it makes a complete round trip along a circular path? Explain with reasoning?
Ans. When the satellite makes a complete round trip along a circular path, it came back to its starting position. Displacement is the change in position of an object from its starting position, so it has not changed its position from its initial position. Hence displacement is 0.
16.Can the displacement be greater than the distance travelled by an object?
Ans. Displacement can not be greater than the distance traveled by an object, because displacement is always less than or equal to the distance traveled, if they are the same direction, then their magnitude will be equal, however, if they have different directions then displacement will be less than the distance.
17. What is negative acceleration?
Ans. An object which moves in the positive direction has a positive velocity and if the speed of the object decreases, it is called negative acceleration or deceleration or retardation.
18. A particle moving with a uniform velocity. What is its acceleration?
Ans. The rate at which the velocity of an object changes, is called the acceleration of the object, so when the particle moving with a uniform velocity its acceleration will be zero.
19. Is it possible for a body to have zero velocity but constant acceleration? Justify your answer.
Ans. Yes, it is possible for a body to have zero velocity but constant acceleration because when a body moving along a circular path changes its direction continuously and its velocity is therefore not constant even if its acceleration is constant.
20. Study the given graph and answer the following questions?
(a) Which part of the graph shows accelerated motion?
(b) Which part of the graph shows retarded motion?
(c) Calculate the distance traveled by the body in the first 4 sec of the journey graphically?
Ans. (a) AB shows accelerated motion.
(b) CD shows retarded motion.
(c) Distance travelled by the body in first 4 sec. = Area of triangle ABE
= 1/2 x AE x BE = 1/2 x 4 x 4
= 8 m.
21. The velocity-time graph of an object is shown in the fig:
(a) State the kind of motion that object has, from A to B and B to C.
(b) Identify the part of the graph where the object has zero acceleration.
Give reasons for your answer.
(c) Identify the part of the graph where the object has negative acceleration. Give reasons for your answer.
Ans. (a) Uniform motion from A to B -Velocity is constant that is 40 m/s.
Non-uniform motion from B to C.
(b) AB has zero acceleration because velocity remains constant from A to B.
(c) BC has negative acceleration because velocity decreases from B to C.
22. The distance moved by a student at different intervals of time, while walking to school, is given in the table below:
Draw the distance-time graph for the motion of the student indicating the scale chosen. What does the shape of the graph suggest about the type of motion?
Time for a starting point(s)
|
0
|
10
|
20
|
30
|
40
|
50
|
Distance moved(m)
|
0
|
15
|
30
|
45
|
60
|
75
|
Ans.
The graph suggests that it is a uniform motion, in which the student covers the equal distance in an equal interval of time.
23. Study the table:
(a) Is the car moving with constant speed?
(b) What is the average speed?
(c) Which duration represents the maximum velocity?
Time
|
10:30am
|
11:00 am
|
11:30 am
|
12:00 noon
|
12:30 pm
|
Distance from origin point.(km)
|
0
|
15
|
28
|
40
|
60
|
(b) Average speed = Total distance /Total time
= (15 + 28+ 40+ 60)/ 12:30 - 10:30 = 143 / 2 = 76.5 m/s.
(c) Between 12:00 noon to 12:30 pm.
24. From displacement time graph shown in fig, find
(a) The velocity of A.
(b) The velocity of B.
Ans. (i) Velocity of A
Displacement = 80 -20 = 60m and time = 8 s.
Velocity = d / t = 60/8 = 7.5 m/s
(ii) Velocity of B
Displacement = 60 - 40 = 20 m
Velocity = d/t = 20/8 = 2.5 m/s.
25. During an experiment, a signal from a spaceship reached the ground station in 5 min. What was the distance of the spaceship from the ground station? The signal travels at the speed of light that is 3 x 10^8 m/s.
Ans. Given: Time = 5 min = 5 x 60 = 300 s
Speed of signal or light = 3 x 10^8 m/s
Distance = speed x time
= 3 x 108 x 300
= 9 x 1010 m
= 9 x 107 km.
26. The bus decreases its speed from 80 km/hr to 60 km/hr in 5 sec. Find the acceleration of the bus?
Ans. Here initial velocity u = 80 km/hr and final velocity v = 60 km/hr.
time= 5 sec.
Acceleration a = v - u /t = 60 - 80 / 5 = -20/5= -6 m/s2 (Retardation)
III. Long Answer type questions:
1. Differentiate between:
(a) Distance and displacement.
(b) Speed and velocity.
(c) Scalar and vectors.
(d) Uniform and non-uniform motion.
Ans. (i) Differences between distance and displacement -
Distance is the length of the path traversed by the object in a certain time but displacement is the shortest distance between the initial and final position of the object.
Distance is a scalar quantity, it has only the magnitude but displacement is a vector quantity and has both the magnitude and direction.
distance is always positive but displacement can be positive, negative or zero.
The distance can be more than or equal to the displacement but displacement can never be greater than the distance, it can be equal or less than distance.
(ii) Difference between speed and velocity -
The rate of motion of an object or the distance traveled by the object in unit time is called speed. The SI unit of speed is meter per second but the velocity is the rate of motion of an object or the distance traveled by the object in a particular direction in unit time is called velocity.
b. Speed is a scalar quantity but velocity is a vector quantity.
c. Speed is always positive but velocity can be positive, negative or zero.
(iii) Uniform and non-uniform motion-
If a moving object covers equal distances in equal intervals of time, it is said to be in uniform motion.
but if the object covers unequal distances in equal intervals of time, it is said to be in non-uniform motion.
2. Answer the following questions:
(i) State the type of motion shown by a freely falling stone.
(ii) When a stone is thrown vertically upwards its velocity is continuously decreased. Why?
(iii) Give an example of a motion in which average velocity is zero, but the average speed is not zero.
Ans. (i) Uniform accelerated motion.
(ii) When a stone is thrown vertically upwards its velocity is continuously decreasing because the earth's gravity pulls it down towards the surface.
(iii) In one rotation in a circular path, the average velocity is zero, but the average speed is not zero.
3. The velocity-time graph shown below represents the motion of a body:
(a) During which interval of time, the body is moving with maximum acceleration?
(b) Calculate the average velocity for the entire journey.
Time in sec.
Ans. (i) In the interval of time 0 - 10 sec (OA) acceleration a = (v - u)/t
= (4 - 0)/10 = 0.4 m/s2
In the interval of time 10 - 20 sec (AB) acceleration a = (v - u)/t
= 5 - 4/20 - 10= 1/10 = 0.1m/s2
In the interval of time 20 - 30 sec (BC) acceleration a = (v - u)/t = 7 - 5/30 - 20 = 2/10 = 0.2 m/s2.
In the interval of time 0 - 10 sec (OA) acceleration is maximum.
or In the 0 - 10 interval of time slop is maximum so at this interval of time acceleration is maximum.
(ii) Distance travelled by body = area of triangle + area of trap 1 + area of trap 2 = 1/2 x 10 x 4 + 1/2 (5 + 4) x 10 + 1/2 (7 + 5) x 10
= 1/2(40 + 90 + 120) = 250/2 = 125 m Ans.
4. A cyclist traveled 3/4 of a circular track from A to B as shown in fig. The radius of the circular track is 400 m.
(a) What is the distance traveled by the cyclist?
(b) What is the displacement?
Ans. (a) Distance travelled by the cyclist = 3/4 of circumference=3/4 x 2πr
= 3/4 x 2 x 22/7 x 400
= 1885.71 m
(b) Displacement AB (shortest distance between A and B)
= ✓(400^2 + 400^2)
=✓[400^2 (1 + 1)]
=400✓2 m Ans.
5. The velocity-time graph shows the motion of a cyclist.
Find
(i) Its acceleration
(ii) Its velocity after 20 s and
(iii)The distance covered by the cyclist in 15 sec. Give reasons for your answer.
Ans. (i) Acceleration is zero because velocity is constant.
(ii) Velocity is constant, so after 20 sec its velocity remains the same that is 20 m/s.
(iii)
Distance covered by the cyclist in 15 sec = Area of ABCD
= AB x BC
= 20 x 15
= 300 m.
6. Derive graphically the first equation of motion:
(i) V = u + at?
(ii) S= Ut + 1/2 at2
(iii) v2 - u2 = 2as.
Ans.EQUATION FOR VELOCITY-TIME RELATION :
Consider the velocity-time graph of an object that moves under uniform acceleration as shown in Fig. From this graph:
The velocity changes at a uniform rate.
In Fig. the initial velocity (u) is represented by OA, the final velocity(v)is represented by BC and the time interval t is represented by OC.
BD = BC – CD (the change in velocity in time interval t)
AD parallel to OC and OC = AD = t
From the graph, we observe that:
BC = BD + DC
= BD + OA
Substituting BC = v and OA = u, we get
v = BD + u or BD = v – u
The acceleration of the object is given by
a = Change in velocity /time taken
= BD = BD = BC - DC
AD OC OC
a = BC - DC
OC
Substituting OC = t, BC = v and DC = u, we get
a = v - u
t
or v - u = at
v = u + at
EQUATION FOR POSITION-TIME RELATION
Let us consider that the object has traveled a distance s in time t under uniform acceleration a. In Fig. 8.8, the distance traveled by the object is obtained by the area enclosed within OABC under the velocity-time graph AB. Thus, the distance s traveled by the object is given by
s = area OABC (which is a trapezium)
= area of the rectangle OADC + area of the triangle ABD
= OA × OC + 1/2 (AD × BD)
s = u × t + 1/2x t×at (We have v = u + at and at = v-u =BD, therefore BD=at)
s = ut + 1/2 at2
EQUATION FOR POSITION–VELOCITY RELATION
From the velocity-time graph shown in fig,the distance s travelled by the object in time t, moving under uniform acceleration a is given by the area enclosed within the trapezium OABC under the graph.
s = area of the trapezium OABC =1/2 (OA + BC) ×OC
Substituting OA = u, BC = v and OC = t, we get
s = 1/2(u+v) t
We have v = u + at
at= v - u
t = v - u
a
s = 1/2 (v+u)(v-u)
a
2as = (v+u)(v-u)= v2 - u2
2as = v2 - u2
7. Study the speed-time graph of a car alongside and answer the following:
(a) What type of motion represented by AB?
(b) Find acceleration from C to D.
(c) Calculate the distance covered by the body from B to C. Give reasons for your answer. Scale of graph (v) : 1 cm = 20 m/s on Y-axis
1 cm = 2 sec on X axis
Ans. (i) AB represents uniform acceleration.
(ii) Acceleration a = (v - u)/t
= (0- 80)/14 - 10
= - 80/4
= -20 m/s2
(iii) The distance covered by body from B to C
= Area of rect. BCFE= BC x CF
= (10 - 4) x 80 [4cm x 20=80]
= 6 x 80 = 480 m
8. A motorboat starting from rest on a lake acceleration in a straight line at a constant rate of 3.0 m/s2 for 8.0 sec. How far does the boat travel during this time?
Ans. Acceleration a = 3.0 m/s2
Time t = 8.0 sec
u = 0
Distance covered by boat s = ut + 1/2 at2
s= 0 x 8 + 1/2 x 3 x 8 x 8
= 96 m
A ball is thrown vertically upwards and returns to the thrower after 12 sec(g=9.8 2 m/s2 )find:
(i) The velocity with which it was thrown up
(ii) The maximum height it reaches.
(iii) Its position after 8 sec.- 16. When will you say a body is in :
(a) Uniform acceleration
(b) Non-uniform acceleration? Draw a velocity –time graph for each type of motion.
9. A train leaves New Delhi railway station at 09:00 AM and reaches Jaipur, which is at a distance of 260 km at 12:45 PM. The train reaches Alwar (at distance 150 km from New Delhi) at 11:30 AM and stops there for 15 minutes.
(i) What is the reference point for the motion of the train?
(ii) What is the average speed to the train between Alwar and Jaipur?
Ans. (i) The reference point is Delhi as all distances and time are measured from there.
(ii) Distance between Alwar and Jaipur = 260-150 = 110 km
Train starts from Alwar at 11:45( 11:30 + 15 = 11:45) as it stops for 15 minutes.
Therefore, time taken by the train to reach Jaipur from alwar = 12:45 - 11:45 = 1 hour
Average speed = Total distance/Total Time = 110/1 = 110 Km/h
10. A scooter starts from rest, moves in a straight line with a constant acceleration and covers a distance of 64 m in 4 sec.
(a) Calculate its acceleration and its final velocity.
(b) In how much time the scooter will cover half the total distance?
Ans. Since scooter starts from rest, Initial velocity = 0
Distance covered by scooter = 64 m and time is taken = 4 s.
Distance S = ut + 1/2 at2
64 = 1/2 x a x 4 x 4
64 = 8 a
Therefore (a) = 64 / 8 = 8 m/s2.
Speed v = u + at = 0 + 8 x 4 = 32 m/s
(a) Acceleration = 8 m/s2 and final velocity =32 m/s.
Total distance = 64 m and half distance = 32 m.
(b) Time taken by the scooter to cover half distance = Distance/Speed
= 32/32
=1 sec.
11. A body covers one complete around a circular park of circumference 176 m in 4 min. Find the displacement of the body after 6 min.
Ans. Distance is the path traveled by a body and displacement is the shortest distance between the initial and final position.
Given that circumference (2πR) = 176 m.
2πR = 176 m
R = 176/2π
= 176x7
2x22
R = 28m
Given 4 min = 1 revolution
Therefore 6 min = 1/4 x 6 = 1 and half revolution.
1 revolution = displacement = 0.
and in half revolution,displacement = diameter = 2R = 2 x 28m = 56m .
12. (a)When will you say a body is in
(i) Uniform acceleration
(ii) Non-uniform acceleration.
(b) A train starts from rest and accelerates uniformly for 30 s to acquire a velocity of 108 km/h. It travels with this velocity for 20 min. The driver now applies breaks and the train retards uniformly to stop after 20 s. Find the total distance covered by the train.
Ans. (a) (i) A body is in uniform acceleration when it travels in a straight path when its velocity increases or decreases by an equal amount with an equal interval of time.
(ii) A body is in nonuniform acceleration when it travels in a straight path when its velocity increases or decreases with the unequal amount with an interval of time.
(b) Here u = 0, t = 30 s, v = 108 km/h = 108 x 5/18 = 30 m/s.
Now acceleration a = v - u /t = (30 - 0)/30 = 1 m/s2
Distance 1 = ut + 1/2 at2
= 0 x 30 + 1/2 x 1 x 30 x 30
= 450 m.
At uniform velocity of 30 m/s for 20 min = 20 x 60 = 1200 s.
Distance 2 = s x t = 30 x 1200 = 36000 m.
Acceleration a = (v - u)/t = (0 - 30)/20 = -3/2=- 1.5 m/s2
Distance 3 = (v2-u2)/2a
= (0 - 302)/2 x 1.5
= 300 m
Total distance = D1 + D2 + D3 = 450 + 36000 + 300
= 36750 m
13. Derive mathematically the first equation of motion:
(i) V = u + at?
(ii) S= Ut + 1/2 at2
(iii) v2 - u2 = 2as.
Ans. (i) Let the time be t, Initial velocity u and final velocity v of a moving body.
We know that acceleration is the rate of change in motion.
a = final velocity - initial velocity / time = v - u /t
at = v - u
v = u + at ..... equation (i)
(ii) We know that Average velocity = Total distance / Total time
= Initial velocity + final velocity
2
= u + v
2
Distance s = Av. velocity x time = (u + v)/2 x t
2s = (v + u) x t
2s = (u + at + u) x t [ v = u + at ]
2s = 2ut + at2
s = ut + 1/2at2..... equation (ii)
(iii) We know that Average velocity = Total distance / Total time
= Initial velocity + final velocity
2
= u + v
2
Distance s = Av. velocity x time = (u + v)/2 x t
2s = (v + u) x t
2s = (v+ u) x (v - u)/a [ v = u + at and t = v - u/a]
2s = v2 - u2 /a
2as = v2 - u2 ..... equation (iii)
13. Define:
(a) Displacement (b) Speed. (c) Uniform speed and non-uniform speed (d) Average speed (e) Velocity (f) Uniform velocity (g)Average velocity (h) Instantaneous velocity (i) Acceleration (j) Uniform acceleration and nonuniform acceleration
(k) Retardation ( deceleration)
Ans. (a) Displacement -It is the shortest distance between the initial and final positions traveled by the object. It is a vector quantity. S.I unit is m.
Define rest and motion.
(b) Speed - The distance covered by a body in a unit interval of time is called speed. It is a scalar quantity. S.I unit is m/s
speed = Distance / Time
(c) Uniform and nonuniform speed- If a body covers equal distance in equal interval of time throughout its motion is said to be uniform speed and if If a body covers the unequal distance in equal interval of time throughout its motion is said to be nonuniform speed.
(d) Average speed- The ratio of the total distance traveled by the body to the total time of the journey is called average speed.
Av. Speed = Total distance / Total time.
(e) Velocity - The distance traveled by a body in a specified direction is called velocity. It is a vector quantity. S.I unit is m/s.
Velocity = Displacement / Time.
(f). Uniform velocity and non-uniform velocity- If a body covers equal distance in equal interval of time throughout its motion along a particular direction, is said to be uniform velocity and if If a body covers the unequal distance in equal interval of time throughout its motion in a particular direction is said to be non-uniform velocity.
(g) Average velocity - The ratio of the total displacement traveled by the body to the total time of the journey is called average velocity.
Av. velocity = Total displacement / Total time.
(h) Instantaneous velocity - When a body moving with a variable velocity, the velocity of the body at any instant is called instantaneous velocity.
(i) Acceleration - The rate of change of velocity with time is called the acceleration.S.I unit is m/s^2.
Acceleration (a) = Change in velocity/time.
(j) Uniform acceleration and nonuniform acceleration- When the equal change in velocity takes place in an equal interval of time, the acceleration is called uniform acceleration.
When an unequal change in velocity takes place in an equal interval of time, the acceleration is called non-uniform acceleration.
(k) Retardation ( deceleration)- If the velocity of a body decreases with time, the motion is said to be declaration or retardation.
NCERT QUESTIONS
- An object has moved through a distance. Can it have zero displacements? If yes, support your answer with an example.
- A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?
- Which of the following is true for displacement?
(a) It cannot be zero. (b) Its magnitude is greater than the distance traveled by the object. - Distinguish between speed and velocity.
- Under what condition(s) is the magnitude of the average velocity of an object equal to its average speed?
- What does the odometer of an automobile measure?
- What does the path of an object look like when it is in uniform motion?
- During an experiment, a signal from a spaceship reached the 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, 3 × 108m s–1.
- When will you say a body is in(i) uniform acceleration? (ii) non-uniform acceleration?
- A bus decreases its speed from 80 km h–1 to 60 km h–1 in 5 s. Find the acceleration of the bus.
- A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km h–1 in 10 minutes. Find its acceleration.
- What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
- What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
- What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
- What is the quantity which is measured by the area occupied below the velocity-time graph?
- 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 traveled.
- 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.
- A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2. What will be its velocity 3 s after the start?
- A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start?
- 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?
- 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?
- Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 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? - 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 40 km h–1. What is the average speed for Abdul’s trip?
- 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?
- A driver of a car traveling 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 traveled farther after the brakes were applied?
- 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?
- 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 in a certain direction with an acceleration in the perpendicular direction. - 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.
No comments:
Post a Comment