# Question-1

Object has moved through a distance, can it have zero displacement? If yes, support your answer with an example.

Solution:
Yes, suppose a person throws a ball upwards through height ‘h’ and catches back the ball. Then, the distance covered by the ball = h + h = 2h. Displacement of the ball= 0.

# Question-2

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 minute 20 seconds?

Solution:
If the farmer starts from point A, then at the end of 2 minutes and 20 seconds (=140 seconds), he will reach the diagonally opposite corner C .The magnitude of displacement of the farmer is:

AC= = 10
2 = 10× 1.414 = 14.14m.

# Question-3

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.

Solution:
Neither (a) nor (b) is true

# Question-4

Distinguish between speed and velocity.

Solution:
Speed of a body needs only magnitude and not direction for its complete description , while the velocity needs both magnitude and direction for its complete description. So speed is a scalar quantity. For example, when a car moves with a speed of 60Km/h in the direction 45° south of west, we can say that the car moves with a velocity of 60km/h 45° south of west. Here the word velocity replaces both the words speed and direction.

# Question-5

Under what condition is the magnitude of average velocity of an object equal to its average speed?

Solution:
Average speed=

Average velocity =

When an object moves along a straight line in the same direction, its total path length is the magnitude of displacement. Hence its average speed is equal to the magnitude of average velocity.

# Question-6

What does the odometer of an automobile measure?

Solution:
The odometer of an automobile measure the distance moved by it.

# Question-7

What does the path of an object look like when it is in uniform motion?

Solution:
The path is a straight line.

# Question-8

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 i.e., 3 x 108m/s.

Solution:
Here t = 5 minutes = 300 s, v = 3 × 108m/s

Distance of spaceship, s = v
× t = 3× 108× 300=9× 1010m.

# Question-9

When will you say a body is in (i) uniform acceleration, (ii) non-uniform acceleration?

Solution:
(i) If a body travels in a straight line and its velocity changes by equal amounts in equal intervals of time, however small these time intervals may be, then the body is said to be in uniform acceleration.

(ii) If the velocity of a body changes by unequal amounts in equal intervals of time, then the body is said to be in non-uniform acceleration.

# Question-10

A bus decreases its speed from 80km/h to 60km/h in 5 seconds. Find the acceleration of the bus.

Solution:
Initial speed, u = 80= = ms-1

Final speed, v = 60==ms-1

Acceleration, a==
ms-2

=-ms-2=-1.11ms-2.

# Question-11

A train starting from the railway station and moving with a uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.

Solution:
Initial speed, u = 0 m/s
Final speed, v = 40===ms
-1

Time, t=10 min=600s
Acceleration, a===ms
-2

=
ms-2

# Question-12

A bus starting from rest moves with a uniform acceleration of 0.1ms-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Solution:
Here, u = 0, a = 0.1 ms-2, t = 2 min=120s

(a) v = u+at =0 + 0.1x 120 =12 ms
-1

(b) s = ut + at2 = 0× 120+× 0.1 × (120)2 = 720 m.

# Question-13

A train is travelling at a speed of 90 km per hour. Brakes are applied so as to produce a uniform acceleration of –0.5ms-2. Find how far the train will go before it is brought to rest?

Solution:
Here, u = 90 km/h = 90 × ms-1 = 25 ms-1

a = -0.5 ms-2, v = 0

As v2 – u2 = 2as

02 – 252 = 2 × (-0.5) × s

or s = = 625 m.

# Question-14

A trolley while going down an inclined plane has an acceleration of 2cms-2. What will be its velocity 3 seconds after the start?

Solution:
Here, u = 0, a = 2 cm s-2, t = 3 s

v = u + at = 0 + 2
× 3 = 6 cm s-1.

# Question-15

A racing car has a uniform acceleration of 4ms-2. What distance will it cover in 10 seconds after start?

Solution:
Here, u = 0, a = 4 m s-2, t = 10 s

s = ut + at2 = 0
× 10 + × 4 × (10)2 = 200 m.

# Question-16

A stone is thrown in a vertically upward direction with a velocity of 5ms-1. If the
acceleration of the stone during its motion is 10 ms-2 in downward direction, what will be the height attained by the stone how much time it will take to reach there?

Solution:
Here, u = 5 ms-1

As the acceleration acts in the opposite direction of initial velocity, so it is negative.

a = -10 ms
-2

At the highest point, v = 0

Using, v2 – u2 = 2as, we get

02 – 52 = 2
× (-10) × s

s = = 1.25 m

Height attained by the stone = 1.25 m

Again, v = u + at

0 = 5 – 10
× t

or t = = 0.5 s.

Time taken by the stone to reach the highest point = 0.5 s.

# Question-17

Ravi told his friend that his house is 1 km towards south from the Main Post Office. Express displacement and the distance moved by the friend from the Post Office when he arrives at Ravi’s house. Mention the reference point chosen by you.

Solution:

Let the reference point chosen be O.

Displacement = 1 km

Distance = 1 km.

# Question-18

In a long distance race, the athlete were expected to take four rounds of the track such that the line of finish was same as the line of start. Suppose the length of the track was 200 m.

(a) What is the total distance to be covered by the athletes?

(b) What is the displacement of the athletes when they touch the finish line?

(c) Is the motion of the athletes’ uniform or non-uniform?

(d) Is the displacement of an athlete and the distance moved by him at the end of the race equal?

Solution:
(a) Distance covered = 200 ´ 4 = 800 m

(b) Zero

(c) Non-uniform motion

(d) Unequal.

# Question-19

The velocity of a car is 18 m/s. Express this velocity in km/h.

Solution:
Velocity of the car = 18 m/s

=
= 64.8 km/h.

# Question-20

Suppose you walk across a room of length 9m with a velocity of one and half kilometer per hour. Express this velocity in unit of m/s and find the time you will take to move across the room.

Solution:
Velocity = km/h

= = 0.42 m/s

0.42m distance is covered in 1 second.

Therefore time taken to cover a distance of 9m is

= = 21.6 s.

# Question-21

What is meant by uniform motion? Can you think of an example of a body in uniform motion?

Solution:
A body is said to have uniform motion when it travels equal distances in equal intervals of time.
For example, a body moving with a constant velocity is said to be in uniform motion.

# Question-22

An electric train is moving with a velocity of 120 km/h. How much distance will it move in 30s?

Solution:
Velocity of the electric train = 120 km/h
=

Hence distance travelled in 1 second = m/s

Distance travelled in 30 second = 30
´ = 1000 m

Hence the train will move 1000 m in 30 seconds.

# Question-23

A body is moving with a velocity of 10 m/s. If the motion is uniform, what will be the velocity after 10s.

Solution:
If the motion of a body is a uniform motion then the body will be travelling with the same velocity. Hence the velocity of the body after 10s will be 10 m/s.

# Question-24

A body moves with a velocity of 2m/s for 5s then its velocity uniformly increases to 10m/s in next 5s. Therefore its velocity begins to decrease at a uniform rate until it comes to rest after 10s.

(a) Plot a velocity-time and distance-time graph for the motion of the body.

(b) Mark the portions of the graph to show when the motion of the body is uniform and when it is non-uniform.

(c) From the graph find the total distance moved by the body after 2s and 12 s and in the last 10s.

Solution:

(b) Uniform motion BC and CD

Non-uniform motion OB

(c) Distance moved after 2s = 4m

Distance moved after 12s = 45 m

Distance moved in last 10s = m

= 10 m.

# Question-25

A cheetah is the fastest land animal and can achieve a peak velocity of 100 km/h up to distances less than 500 m. If a cheetah spots his prey at a distance of 100 m, what is the minimum time it will take to get its prey, if the average velocity attained by it is 90 km/h.

Solution:
Peak velocity of the cheetah = 100 km/h

= m

m distance is covered by cheetah in 1s.

Minimum time required by the cheetah to cover a distance of 100 m is

= = 3.6 s

Therefore the minimum time the cheetah will take to get its prey = 3.6 s.

# Question-26

All buses and cars these days are fitted with a speedometer, which shows the velocity of the vehicle. A device called odometer records the distance moved by the vehicle. If the reading on the odometer of a vehicle in the beginning of a trip and after 40 minutes were 1048 km and 1096 km respectively, calculate its average velocity. Will the reading on the speedometer show this velocity when the vehicle is moving? Support your answer with reason.

Solution:
Total distance travelled = (1096 – 1048) km = 48 km

Total time taken = 40 minutes

Average speed =

= = 1.2 km/min = 20m/s

The reading of the speedometer will not show this velocity because when the road is straight flat and free, the speed may be much more than 20 m/s, but on curved roads, hills or in a crowded area, the speed may fall below this average value (20 m/s).

# Question-27

Two cars moving in opposite directions cover same distance in one hour. If the cars were moving in north-south direction, what will be their displacement in one hour?

Solution:
When the cars move in north-south direction their displacement in one hour will be ‘zero’.

# Question-28

A sprinter in a 100 m race, covers 4m in first second, 30 m in next 4s, 52 m in another 4s and finishes the race in 10s.
(a) Calculate the average velocity of the sprinter.

(b) In which time interval, is the average velocity attained by the sprinter maximum? State this velocity in appropriate units.

(c) Plot the distance-time graph for the motion of the sprinter.

(d) Find out the distance moved by the sprinter at the end of 6s with the help of the graph.

Solution:
(a) Average velocity of the sprinter =

= = 10 m/s
(b) (i) v1 = = 4 m/s

(ii) v2 = = 7.5 m/s

(iii) v3 = = 13 m/s

(iv) v4 = = 14 m/s
Maximum velocity attained by sprinter = 14 m/s.

Hence in the last one second the sprinter attained maximum velocity.

(c)

(d)

Velocity of the sprinter at the end of 6s

= = 7.8 m/s.

# Question-29

A stone is thrown in vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion be 10 m/s2, in downward direction what will be height attained by the stone and how much time will it take to reach there?

Solution:
Initial velocity u = 5 m/s

Acceleration a = 10 m/s2

We know that

v2 = u2 + 2gh

0 = (5)2 + 2 × (-9.8) × h

= 25 - 19.6h

h = = 1.3 m

Also v = u + gt

0 = 5 – 9.8t

or     t = = 0.5 s.

# Question-30

State which of the following situations are possible and give an example of each of these:

(a) a body moving with constant acceleration but with zero velocity.

(b) A body moving horizontally with an acceleration in vertical direction.

(c) A body moving with a constant velocity in an accelerated motion.

Solution:
All the situations stated above are possible.

(a) When a body is thrown vertically upward in space, then at the highest point, the body has zero velocity but the acceleration of the body will be equal to the acceleration due to gravity.

(b) A body moving horizontally with an acceleration in vertical direction is possible in a projectile motion.

(c) A body moving with a constant velocity in an accelerated motion is possible in a uniform circular motion.

# Question-31

The average time taken by a normal person to react to an emergency is one fifteenth of a second and is called the ‘reaction time’. If a bus is moving with a velocity of 60 km/h and its driver sees a child running across the road, how much distance would that bus had moved before he could press the brakes? The reaction time of the people increases when they are intoxicated. How much distance had the bus moved if the reaction time of the driver were ½ s under the influence of alcohol?

Solution:
Initial velocity u = 60 km/h

= m/s

Reaction time = s

Distance travelled in 1s = m

Distance travelled in s = = 1.1 m

Reaction time of the driver = s

Distance travelled in 1 s = m

Distance travelled in s = = 8.3 m
.

# Math Notes

We’ll discuss many of the concepts in this chapter in depth later. But for now, we need a brief review of these concepts for many of the problems that follow.

1. To compare two fractions, cross-multiply. The larger product will be on the same side as the larger fraction.

Example: Given 5/6 vs. 6/7. Cross-multiplying gives 5 7 vs. 6 6, or 35 vs. 36. Now 36 is larger than 35, so 6/7 is larger than 5/6.

2. Taking the square root of a fraction between 0 and 1 makes it larger.

Example:  and 1/2 is greater than 1/4.

Caution: This is not true for fractions greater than 1. For example, . But .

3. Squaring a fraction between 0 and 1 makes it smaller.

Example:  and 1/4 is less than 1/2.

4. ax2 ≠ (ax)2. In fact, a2x2 = (ax)2.

Example: 3 22 = 3 4 = 12. But (3 2)2 = 62 = 36. This mistake is often seen in the following form: –x2 = (–x)2. To see more clearly why this is wrong, write –x2 = (–1)x2, which is negative. But (–x)2 = (–x)(–x) = x2, which is positive.

Example: –52 = (–1)52 = (–1)25 = –25.  But (–5)2 = (–5)(–5) = 5 5 = 25.

5. . In fact,  and .

Example:  . But .

6. –(a + b) ≠ –a + b. In fact, –(a + b) = –a – b.

Example:  –(2 + 3) = –5. But –2 + 3 = 1.

Example:  –(2 + x) = –2 – x.

7. Memorize the following factoring formulas—they occur frequently on the test.

1. x2y2 = (x + y)(xy)
2. x2 ± 2xy + y2 = (x ± y)2
3. a(b + c) = ab + ac
8. Know these rules for radicals:
9. Pythagorean Theorem (For right triangles only):
c2 = a2 + b2

Example:

 Column A Column B 10 The area of the triangle

Solution: Since the triangle is a right triangle, the Pythagorean Theorem applies:
h2 + 32 = 52, where h is the height of the triangle. Solving for h yields h = 4. Hence, the area of the triangle is  .

10. When parallel lines are cut by a transversal, three important angle relationships are formed:

11. In a triangle, an exterior angle is equal to the sum of its remote interior angles and therefore greater than either of them.

e = a + b  and e > a  and e > b

12. A central angle has by definition the same measure as its intercepted arc.

13. An inscribed angle has one-half the measure of its intercepted arc.

14. There are 180° in a straight angle.

15. The angle sum of a triangle is 180°.

Example:

 Column A Column B 30 The degree measure of angle c

Solution: Since a triangle has 180˚, we get 100 + 50 + c = 180. Solving for c yields c = 30. Hence, the columns are equal, and the answer is (C).

16. To find the percentage increase, find the absolute increase and divide by the original amount.

Example: If a shirt selling for $18 is marked up to$20, then the absolute increase is 20 – 18 = 2. Thus, the percentage increase is

17. Systems of simultaneous equations can most often be solved by merely adding or subtracting the equations.

Example: If 4x + y = 14 and 3x + 2y = 13, then x – y =

Solution: Merely subtract the second equation from the first:

18. Rounding Off:

The convention used for rounding numbers is “if the following digit is less than five, then the preceding digit is not changed. But if the following digit is greater than or equal to five, then the preceding digit is increased by one.”

Example: 65,439 —> 65,000 (following digit is 4)

5.5671 —> 5.5700 (dropping the unnecessary zeros gives 5.57

# Question-33

Discuss the graphs A, B and c shown in the figure. Compare the total distance travelled and the displacements. Which graph represents a motion with negative acceleration?

Solution:
(i) Graph A is distance-time graph

Displacement = 8 m

Distance = 2
×

= 2
×

= 2
× 17 = 34 m

(ii) Graph B is distance-time graph

Displacement = Zero

Distance = = 60 m

(iii) Graph c is a distance-time graph

Displacement = 10 m

Distance = =

= = 17

(iv) Graph C represents a motion with negative acceleration.

# Question-34

An artificial satellite is moving in a circular orbit of radius nearly 42,250 km. Calculate its linear velocity, if it takes 24 hours to revolve round the earth.

Solution:
Time taken t = 24 hours

= 24
× 60 × 60 = 86400 s

Angle subtended at the centre of the earth = 2

Linear velocity v = r
ω

where ω is the angular velocity

Angular velocity =

= 7.27

Linear velocity v = rω = 42,250 × 7.27 × 10-5

= 3.07 km/s.

# Question-35

A circular cycle track has a circumference of 314 m with AB as one of its diameter. A cyclist travels from A to B along the circular path with a velocity of constant magnitude 15.7 m/s/ Fine:
(a) the distance moved by the cyclist.

(b) the displacement of the cyclist if AB represents north-south direction.

(c) the average velocity of the cyclist

Solution:
Diameter of the circular cycle track = AB

Circumference of the cycle track = 314 m
Velocity = 15.7 m/s

(a) Distance travelled by the cyclist = 2
πr
= 314 m

(b) Displacement = 0, since the final position coincides with the initial position.

(c) Average velocity =

15.7 m distance is covered in 1s

Time taken to cover a distance of 314 m is

= = 20 s

Average velocity = = 15.7 m/s

(d) Average acceleration = 0.

# Question-36

A train is travelling at a speed of 90 km/h, the brakes are applied so as to produce a uniform acceleration of –0.5 ms-2. Find how far the train goes before it stops?

Solution:
Initial velocity u = 90 km/h

= = 25 m/s

Acceleration = -0.5 ms-2

Final velocity v = 0

We know that v2 = u2 + 2as

s =

Hence the distance travelled by the train before it stops is 625m.