Question1
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.
Question2
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.
Question3
(a) It cannot be Zero.
(b) Its magnitude is greater than the distance travelled by the object.
Solution:
Neither (a) nor (b) is true
Question4
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.
Question5
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.
Question6
Solution:
The odometer of an automobile measure the distance moved by it.
Question7
Solution:
The path is a straight line.
Question8
Solution:
Here t = 5 minutes = 300 s, v = 3 × 10^{8}m/s
Distance of spaceship, s = v× t = 3× 10^{8}× 300=9× 10^{10}m.
Question9
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 nonuniform acceleration.
Question10
Solution:
Initial speed, u = 80= = ms^{1}
Final speed, v = 60==ms^{1 Acceleration, a== }ms^{2}
=ms^{2}=1.11ms^{2}.
Question11
Solution:
Initial speed, u = 0 m/s
Final speed, v = 40===ms^{1}
Time, t=10 min=600s
Acceleration, a===ms^{2 =}ms^{2}
Question12
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 + at^{2} = 0× 120+× 0.1 × (120)^{2 }= 720 m.
Question13
Solution:
Here, u = 90 km/h = 90 × ms^{1} = 25 ms^{1}
a = 0.5 ms^{2}, v = 0
As v^{2} – u^{2} = 2as
∴ 0^{2} – 25^{2} = 2 × (0.5) × s
or s = = 625 m.
Question14
Solution:
Here, u = 0, a = 2 cm s^{2}, t = 3 s
v = u + at = 0 + 2 × 3 = 6 cm s^{1}.
Question15
Solution:
Here, u = 0, a = 4 m s^{2}, t = 10 s
s = ut + at^{2} = 0 × 10 + × 4 × (10)^{2} = 200 m.
Question16
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, v^{2} – u^{2} = 2as, we get
0^{2} – 5^{2} = 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.
Question17
Solution:
Let the reference point chosen be O.
Displacement = 1 km
Distance = 1 km.
Question18
(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 nonuniform?
(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) Nonuniform motion
(d) Unequal.
Question19
Solution:
Velocity of the car = 18 m/s
=
= 64.8 km/h.
Question20
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.
Question21
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.
Question22
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.
Question23
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.
Question24
(a) Plot a velocitytime and distancetime 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 nonuniform.
(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
Nonuniform motion OB
(c) Distance moved after 2s = 4m
Distance moved after 12s = 45 m
Distance moved in last 10s = m
= 10 m.
Question25
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.
Question26
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).
Question27
Solution:
When the cars move in northsouth direction their displacement in one hour will be ‘zero’.
Question28
(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 distancetime 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) v_{1} = = 4 m/s
(ii) v_{2} = = 7.5 m/s
(iii) v_{3} = = 13 m/s
(iv) v_{4} = = 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.
Question29
Solution:
Initial velocity u = 5 m/s
Acceleration a = 10 m/s^{2}
We know that
v^{2} = u^{2} + 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.
Question30
(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.
Question31
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.
Example: Given 5/6 vs. 6/7. Crossmultiplying gives 5 — 7 vs. 6 — 6, or 35 vs. 36. Now 36 is larger than 35, so 6/7 is larger than 5/6.
Example: and 1/2 is greater than 1/4.
Caution: This is not true for fractions greater than 1. For example, . But .
Example: and 1/4 is less than 1/2.
Example: 3 — 2^{2} = 3 — 4 = 12. But (3 — 2)^{2} = 6^{2} = 36. This mistake is often seen in the following form: –x^{2} = (–x)^{2}. To see more clearly why this is wrong, write –x^{2} = (–1)x^{2}, which is negative. But (–x)^{2} = (–x)(–x) = x^{2}, which is positive.
Example: –5^{2} = (–1)5^{2} = (–1)25 = –25. But (–5)^{2} = (–5)(–5) = 5 — 5 = 25.
Example: . But .
Example: –(2 + 3) = –5. But –2 + 3 = 1.
Example: –(2 + x) = –2 – x.
 x^{2} – y^{2} = (x + y)(x – y)
 x^{2} ± 2xy + y^{2} = (x ± y)^{2}
 a(b + c) = ab + ac
c^{2} = a^{2} + b^{2}
Example:
Column A 
Column B 

10 
The area of the triangle 
Solution: Since the triangle is a right triangle, the Pythagorean Theorem applies: h^{2} + 3^{2} = 5^{2}, where h is the height of the triangle. Solving for h yields h = 4. Hence, the area of the triangle is .
The answer is (A).
e = a + b and e > a and e > b
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).
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
Example: If 4x + y = 14 and 3x + 2y = 13, then x – y =
Solution: Merely subtract the second equation from the first:
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
Question33
Solution:
(i) Graph A is distancetime graph
Displacement = 8 m
Distance = 2 ×
= 2 ×
= 2 × 17 = 34 m
(ii) Graph B is distancetime graph
Displacement = Zero
Distance = = 60 m
(iii) Graph c is a distancetime graph
Displacement = 10 m
Distance = =
= = 17
(iv) Graph C represents a motion with negative acceleration.
Question34
Solution:
Time taken t = 24 hours
= 24 × 60 × 60 = 86400 s
Angle subtended at the centre of the earth = 2π radians
Linear velocity v = rω
where ω is the angular velocity
Angular velocity =
= rad/s
= 7.27 × 10^{5} rad/s
∴ Linear velocity v = rω = 42,250 × 7.27 × 10^{5}
= 3.07 km/s.
Question35
(a) the distance moved by the cyclist.
(b) the displacement of the cyclist if AB represents northsouth 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 = 2πr
= 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.
Question36
Solution:
Initial velocity u = 90 km/h
= = 25 m/s
Acceleration = 0.5 ms^{2}
Final velocity v = 0
We know that v^{2} = u^{2} + 2as
∴ s =
Hence the distance travelled by the train before it stops is 625m.