INTEXT QUESTIONS
Q1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Ans.
Yes, suppose a person throws a ball upwards through height h and catches the back the ball, Then,
Distance covered by the ball = h + h = 2h
Displacement of the ball = 0.
Similarly, consider a body moving in a circular path of radius r completes one revolution in time t.
Distance covered by the body = 2$\mathrm{\pi}$r
Displacement of the body = 0
Q2. 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?
Ans. The farmer takes 40 s to cover 4 × 10 = 40 m.
In 2 min and 20 s (140 s), he will cover a distance $=\frac{40}{40}\times 140=140m$
Therefore, the farmer completes $=\frac{140}{40}=3.5rounds$
(3 complete rounds and a half round) of the field in 2 min and 20 s.
That means, after 2 min 20 s, the farmer will be at the opposite end of the starting point.
Now, there can be two extreme cases.
Case I: Starting point is a corner point of the field.
In this case, the farmer will be at the diagonally opposite corner of the field after 2 min 20 s.
Therefore, the displacement will be equal to the diagonal of the field.
Hence, the displacement will be $\sqrt{{10}^{2}+{10}^{2}}=14.1m$
Case II: Starting point is the middle point of any side of the field.
In this case the farmer will be at the middle point of the opposite side of the field after 2 min 20 s.
Therefore, the displacement will be equal to the side of the field, i.e., 10 m.
For any other starting point, the displacement will be between 14.1 m and 10 m.
Q3. Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.
Ans.
(a) Not true
Displacement can become zero when the initial and final position of the object is the same.
(b) Not true
Displacement is the shortest measurable distance between the initial and final positions of an object. It cannot be greater than the magnitude of the distance travelled by an object. However, sometimes, it may be equal to the distance travelled by the object.
INTEXT QUESTIONS
Q1. Distinguish between speed and velocity.
Ans.
Speed |
Velocity |
Speed is the distance travelled by an object in a given interval of time. It does not have any direction. |
Velocity is the displacement of an object in a given interval of time. It has a unique direction. |
Speed is given by the relation: $\text{Speed}=\frac{\text{Distance}}{\text{time}}$ |
Velocity is given by the relation: $\text{velocity}=\frac{\text{Displacement}}{\text{time}}$ |
The speed of an object can never be negative. At the most, it can become zero. This is because distance travelled can never be negative. |
The velocity of an object can be negative, positive, or equal to zero. This is because displacement can take any of these three values. |
Q2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Ans.
$Averagespeed=\frac{Totaldis\mathrm{tan}cecovered}{totaltimetaken}$
$Averagevelocity=\frac{Totaldisplacement}{totaltimetaken}$
If the total distance covered by an object is the same as its displacement, then its average speed would be equal to its average velocity.
Q3. What does the odometer of an automobile measure?
Ans.
The odometer of an automobile measures the distance covered by an automobile.
Q4. What does the path of an object look like when it is in uniform motion?
Ans.
An object having uniform motion has a straight line path.
Q5. 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 × 10^{8} m s^{−1}.
Ans.
Time taken by the signal to reach the ground station from the spaceship
= 5 min = 5 × 60 = 300 s
Speed of the signal = 3 × 10^{8} m/s
$Speed=\frac{Dis\mathrm{tan}cecovered}{time}$
$\therefore $Distance travelled = Speed × Time
= 3 × 10^{8} × 300
= 9 × 10^{10} m
Hence, the distance of the spaceship from the ground station is 9 × 10^{10} m.
INTEXT QUESTIONS
Q1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?
Ans.
(i) A body is said to have uniform acceleration if it travels in a straight path in such a way that its velocity changes at a uniform rate, i.e., the velocity of a body increases or decreases by equal amounts in an equal interval of time.
(ii) A body is said to have non-uniform acceleration if it travels in a straight path in such a way that its velocity changes at a non-uniform rate, i.e., the velocity of a body increases or decreases in unequal amounts in an equal interval of time.
Q2. A bus decreases its speed from 80 kmph to 60 km h^{−1} in 5 s. Find the acceleration of the bus.
Ans.
Initial speed of the bus, $u=80km/h=80\times \frac{5}{18}=22.22m/s$
Final speed of the bus, $v=60km/h=60\times \frac{5}{18}=16.66m/s$
Time take to decrease the speed, t = 5 s
$Acceleration=\frac{v\u2013u}{t}$
$\frac{16.66\u201322.22}{5}=\u20131.112m/{s}^{2}$
Here, the negative sign of acceleration indicates that the velocity of the car is decreasing.
Q3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h^{−1} in 10 minutes. Find its acceleration.
Ans.
Initial velocity of the train, u = 0 (since the train is initially at rest)
Final velocity of the train, $v=40km/h=40\times \frac{5}{18}=11.11m/s$
Time taken, t = 10 min=10 × 60 = 600 s
$Acceleration=\frac{v\u2013u}{t}$$=\frac{11.11\u20130}{600}=0.0185m/{s}^{2}$
Hence, the acceleration of the train is 0.0185 m/s^{2}.
INTEXT QUESTIONS
Q1. What is the nature of the distance−time graphs for uniform and non-uniform motion of an object?
Ans.
The distance−time graph for uniform motion of an object is a straight line (as shown in the following figure).
The distance−time graph for non-uniform motion of an object is a curved line (as shown in the given figure).
Q2. What can you say about the motion of an object whose distance−time graph is a straight line parallel to the time axis?
Ans.
When an object is at rest, its distance−time graph is a straight line parallel to the time axis.
A straight line parallel to the x-axis in a distance−-time graph indicates that with a change in time, there is no change in the position of the object. Thus, the object is at rest.
Q3. 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.
Object is moving uniformly.
A straight line parallel to the time axis in a speed−time graph indicates that with a change in time, there is no change in the speed of the object. This indicates the uniform motion of the object.
Q4. What is the quantity which is measured by the area occupied below the velocity−time graph?
Ans.
Distance
The graph shows the velocity−time graph of a uniformly moving body.
Let the velocity of the body at time (t) be v.
Area of the shaded region = length × breath
Where,
Length = t
Breath = v
Area = vt = velocity × time …(i)
We know,
Velocity = Displacement / time
$\therefore $Displacement = Velocity × Time…(ii)
From equations (i) and (ii),
Area = Displacement
Hence, the area occupied below the velocity−time graph measures the displacement covered by the body.
INTEXT QUESTIONS
Q1. 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.
(a) 12 m/s
(b) 720 m
(a) Initial speed of the bus, u = 0
(since the bus is initially at rest)
Acceleration, a = 0.1 m/s^{2}
Time taken, t = 2 minutes = 120 s
Let v be the final speed acquired by the bus.
$\therefore Acceleration\left(a\right)=\frac{v\u2013u}{t}$
$0.1=\frac{v\u20130}{120}$
$\therefore v=12m/s$
(b) According to the third equation of motion:
v^{2} − u^{2} = 2as
Where, s is the distance covered by the bus
(12)^{2} − (0)^{2} = 2(0.1) s
s = 720 m
Speed acquired by the bus is 12 m/s.
Distance travelled by the bus is 720 m.
Q2. 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.
Initial speed of the train, u = 90 km/h = 25 m/s
Final speed of the train, v = 0
finally the train comes to rest)
Acceleration = − 0.5 m s^{−2}
According to third equation of motion:
v^{2} = u^{2} + 2 as
(0)^{2} = (25)^{2} + 2 (− 0.5) s
Where, s is the distance covered by the train
$s=\frac{{\left(25\right)}^{2}}{2\left(0.5\right)}=625m$
The train will cover a distance of 625 m before it comes to rest.
Q3. 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?
Ans.
Initial velocity of the trolley, u = 0
(since the trolley was initially at rest)
Acceleration, a = 2 cm s^{−2} = 0.02 m/s^{2}
Time, t = 3 s
According to the first equation of motion:
v = u + at
Where, v is the velocity of the trolley after 3 s from start
v = 0 + 0.02 × 3 = 0.06 m/s
Hence, the velocity of the trolley after 3 s from start is 0.06 m/s.
Q4. A racing car has a uniform acceleration of 4 m s^{−2}. What distance will it cover in 10 s after start?
Ans.
Initial velocity of the racing car, u = 0 (since the racing car is initially at rest)
Acceleration, a = 4 m/s^{2}
Time taken, t = 10 s
According to the second equation of motion: $s=ut+\frac{1}{2}a{t}^{2}$
Where, s is the distance covered by the racing car $s=0+\frac{1}{2}\times 4\times {\left(10\right)}^{2}=\frac{400}{2}=200m$
Hence, the distance covered by the racing car after 10 s from start is 200 m.
Q5. 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.
Initially, velocity of the stone,u = 5 m/s
Final velocity, v = 0 (since the stone comes to rest when it reaches its maximum height)
Acceleration of the stone, a = acceleration due to gravity, g = 10 m/s^{2}
(in downward direction)
There will be a change in the sign of acceleration because the stone is being thrown upwards.
Acceleration, a = −10 m/s^{2}
Let s be the maximum height attained by the stone in time t.
According to the first equation of motion:
$v=u+at$
$0=5+\left(\u201310\right)t$
$\therefore t=\frac{\u20135}{\u201310}=0.5s$
According to the third equation of motion:
${v}^{2}={u}^{2}+2as$
${\left(0\right)}^{2}={\left(5\right)}^{2}+2\left(\u201310\right)s$
$s=\frac{{5}^{2}}{20}=1.25m$
Hence, the stone attains a height of 1.25 m in 0.5 s.
NCERT TEXT BOOK EXERCISES
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.
Diameter of a circular track, d = 200 m
Radius of the track, $r=\frac{d}{2}=100m$
Circumference$=2\mathrm{\pi r}=2\mathrm{\pi}\left(100\right)=200\mathrm{\pi m}$
In 40 s, the given athlete covers a distance of 200π m.
In 1 s, the given athlete covers a $dis\mathrm{tan}ce=\frac{200\mathrm{\pi}}{40}$
The athlete runs for 2 minutes 20 s = 140 s
$\therefore $Total distance covered in 140s $=\frac{200\times 22}{40\times 4}\times 140=2200m$
The athlete covers one round of the circular track in 40 s. This means that after every 40 s, the athlete comes back to his original position. Hence, in 140 s he had completed 3 rounds of the circular track and is taking the fourth round.
He takes 3 rounds in 40 × 3 = 120 s. Thus, after 120 s his displacement is zero.
Then, the net displacement of the athlete is in 20 s only. In this interval of time, he moves at the opposite end of the initial position. Since displacement is equal to the shortest distance between the initial and final position of the athlete, displacement of the athlete will be equal to the diameter of the circular track.
$\therefore $Displacement of the athlete = 200 m
Distance covered by the athlete in 2 min 20 s is 2200 m and his displacement is 200 m.
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.
(a) 2 m/s, 2 m/s
(b) 1.90 m/s, 0.95 m/s
(a) From end A to end B
Distance covered by Joseph while jogging from A to B = 300 m
Time taken to cover that distance = 2 min 30 seconds = 150 s
$Averagevelocity=\frac{Totaldis\mathrm{tan}cecovered}{totaltimetaken}$
Total distance covered = 300 m
Total time taken = 150 s
Average speed = 300/150 = 2 m/s
$Averagevelocity=\frac{Totaldisplacement}{totaltimetaken}$
Displacement = shortest distance between A and B = 300 m
Time interval = 150 s
Average velocity = 300/150 = 2 m/s
The average speed and average velocity of Joseph from A to B are the same and equal to 2 m/s.
(b) From end A to end C
$Averagespeed=\frac{Totaldis\mathrm{tan}cecovered}{totaltimetaken}$
Total distance covered = Distance from A to B + Distance from B to C = 300 + 100 = 400 m
Total time taken = Time taken to travel from A to B + Time taken to travel from B to C = 150 + 60 = 210 s
Average speed = 400/210 = 1.90 m/s
$Averagevelocity=\frac{Totaldisplacement}{totaltimetaken}$
Displacement from A to C
= AC = AB − BC = 300 − 100 = 200 m
Time interval = time taken to travel from A to B + time taken to travel from B to C = 150 + 60 = 210 s
Average velocity = 200210 = 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 40 km h^{−1}. What is the average speed for Abdul’s trip?
Ans.
Case I: While driving to school
Average speed of Abdul’s trip
$Averagespeed=\frac{Totaldis\mathrm{tan}cecovered}{totaltimetaken}$
Total distance = Distance travelled to reach school = d
Let total time taken = t_{1}
$\therefore 20=\frac{d}{{t}_{1}}$
${t}_{1}=\frac{d}{20}........\left(i\right)$
Case II: While returning from school
Total distance = Distance travelled while returning from school = d
Now, total time taken = t_{2}
$\therefore 40=\frac{d}{{t}_{2}}$
${t}_{2}=\frac{d}{40}......\left(ii\right)$
$Averagespeed=\frac{Totaldis\mathrm{tan}cecovered}{totaltimetaken}$
Where,
Total distance covered in the trip = d + d = 2d
Total time taken, t = Time taken to go to school + Time taken to return to school = t_{1} + t_{2}
$\therefore Avg.speed=\frac{2d}{{t}_{1}+{t}_{2}}$
From equations (i) and (ii),
$\therefore Avg.speed=\frac{2d}{{\displaystyle \frac{d}{20}}+{\displaystyle \frac{d}{40}}}=\frac{2}{{\displaystyle \frac{2+1}{40}}}=\frac{80}{3}=26.67m/s$
Hence, the average speed for Abdul’s trip is 26.67 m/s.
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.
Initial velocity, u = 0 (since the motor boat is initially at rest)
Acceleration of the motorboat, a = 3 m/s^{2}
Time taken, t = 8 s
According to the second equation of motion :
$s=ut+\frac{1}{2}a{t}^{2}$
Distance covered by the motorboat, s
$s=0+\frac{1}{2}\times 3\times (8{)}^{2}=96\mathrm{m}$
Hence, the boat travels a distance of 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.
Case A:
Initial speed of the car, u_{1} = 52 km/h
= 14.4 m/s
Time taken to stop the car, t_{1} = 5 s
Final speed of the car becomes zero after 5 s of application of brakes.
Case B:
Initial speed of the car, u_{2} = 3 km/h
= 0.833 m/s ≅ 0.83 m/s
Time taken to stop the car, t_{2} = 10 s
Final speed of the car becomes zero after 10 s of application of brakes.
Plot of the two cars on a speed−time graph is shown in the following figure:
Distance covered by each car is equal to the area under the speed−time graph.
Distance covered in case A, ${\mathrm{s}}_{1}=\frac{1}{2}\times \mathrm{OP}\times \mathrm{OR}=\frac{1}{2}\times 14.4\times 5=36\mathrm{m}$
Distance covered in case B, ${\mathrm{s}}_{2}=\frac{1}{2}\times \mathrm{OS}\times \mathrm{OQ}=\frac{1}{2}\times 0.63\times 10=4.15\mathrm{m}$
Area of ΔOPR > Area of ΔOSQ
Thus, the distance covered in case A is greater than the distance covered in case B.
Hence, the car travelling with a speed of 52 km/h travels farther after brakes were applied.
Q6. The given figure 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) Object B
(b) No
(c) 5.714
(d) 5.143
(a) $\text{Speed}=\frac{\text{Distance}}{\text{time}}$
$\text{Slope of graph}=\frac{y\text{-axis}}{x\text{-axis}}=\frac{\text{Distance}}{\text{Time}}$
$\therefore $Speed = slope of the graph
Since slope of object B is greater than objects A and C, it is travelling the fastest.
(b) All three objects A, B and C never meet at a single point. Thus, they were never at the same point on road.
(c)
On the distance axis :
7 small boxes = 4 km
$\therefore $1 small box = $\frac{4}{7}km$
Initially, object C is 4 blocks away from the origin.
$\therefore $Initial distance of object C from origin = $\frac{16}{7}km$
Distance of object C from origin when B passes A = 8 km
Distance covered by C $=8\u2013\frac{16}{7}=\frac{56\u201316}{7}=\frac{40}{7}=5.714\mathrm{km}$
Hence, C has travelled a distance of 5.714 km when B passes A.
(d)
Distance covered by B at the time it passes C = 9 boxes
$=\frac{4}{7}\times 9=\frac{36}{7}=5.143\mathrm{km}$
Hence, B has travelled a distance of 5.143 km when B passes A.
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.
Distance covered by the ball, s = 20 m
Acceleration, a = 10 m/s^{2}
Initially, velocity, u = 0 (since the ball was initially at rest)
Final velocity of the ball with which it strikes the ground, v
According to the third equation of motion:
${v}^{2}={u}^{2}+2as\phantom{\rule{0ex}{0ex}}{v}^{2}=0+2\left(10\right)\left(20\right)\phantom{\rule{0ex}{0ex}}v=20\mathrm{m}/\mathrm{s}$
According to the first equation of motion: v = u + at
Where,
Time, t taken by the ball to strike the ground is,
20 = 0 + 10 (t)
t = 2 s
Hence, the ball strikes the ground after 2 s with a velocity of 20 m/s.
Q8. The speed-time graph for a car is shown in the following figure.
(a) Find out how far the car travels 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.
The shaded area which is equal to $\frac{1}{2}\times 4\times 2=12m$ represents the distance travelled by the car in the first 4 s.
(b)
The part of the graph in red colour between time 6 s to 10 s represents uniform motion of the car.
Question 9: 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.
Answer :
(a) Possible
When a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s^{2}.
(b) Possible
When a car is moving in a circular track, its acceleration is perpendicular to its 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.
Radius of the circular orbit,
r = 42250 km
Time taken to revolve around the earth, t = 24 h
Speed of a circular moving object, $\mathrm{v}=\frac{2\pi \mathrm{r}}{\mathrm{t}}$
$=\frac{2\times 3.14\times 42250}{24}=1.105\times {10}^{4}\mathrm{km}/\mathrm{hr}=3.069\mathrm{km}/\mathrm{sec}$
Hence, the speed of the artificial satellite is 3.069 km/s.