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Karnataka State Syllabus Class 8 Science Chapter 8 Describing Motion
KSEEB Class 8 Science Describing Motion Textbook Exercise Questions and Answers
I. Four alternatives are given to each of the following incomplete statements questions. Choose the right answer.
Question 1.
Uniform circular motion is called continuously accelerated motion mainly because
(a) the direction of motion changes.
(b) speed changes continuously
(c) velocity remains the same
(d) the direction of motion does not change
Answer:
(a) the direction of motion changes.
Question 2.
A cricketer hits a sixer, the cricket ball moves up with a velocity of 2 ms-1 and falls down. Its Initial velocity while falling down will be
(a) 1 ms-1
(b) 1 ms-2
(c) 0 ms-1
(d) 2 ms-1
Answer:
(c) 0 ms-1
II. Fill in the blanks with suitable words:
Question:
1. S l unit If acceleration is ____________ .
2. Velocity has both speed and ____________ .
3. If an object starts from ‘A’ and comes back to ‘A’, its displacement will be ____________ .
Answer:
1. metre per second2/ms-2
2. direction
3. 0
III. Solve
Question 1.
An object is moving in a circus lar path of radius 3.5 m. If it completes one full cycle, what will be the displacement and what is the distance travelled?
Answer:
radius = 3.5 m.
[initial point] starting point is A and ending point also A.
Therefore, the lea§t distance between the initial point and final point is 0 therefore the displacement = 0
Distance travelled = 2 πr
= 2 x \(\frac{22}{7}\) x 3.5
= 22 meters.
Question 2.
An object changes its velocity from 30 ms-1 to 40 ms-1 in a time interval of 2 seconds. What is its acceleration?
Answer:
Initial velocity U = 30 ms-1
final velocity v = 40 ms-1
time taken t = 2 sec
acceleration = ?
a = \(\frac{v-u}{t}\)
a = \(\frac{40-30}{2}=\frac{10}{2}\) = 5 ms-2
Question 3.
An object at rest starts moving. It covers a distance of 2m in one second. It covers a further distance of 5 m in two seconds in the same direction what is the average velocity and acceleration?
Answer:
distance travelled on 1 sec = 2 mts
distance travelled in 2 sec = 5 mts
total distance travelled = 2 + 5 = 7 mts
Total time taken to cover 7 mts = 1 + 2 = 3 sec
Average Velocity = \(\frac{\text { Total distance travelled }}{\text { Total time taken }}\)
= \(\frac{7}{3}\) 2.3 ms-1
Average velocity = 2.3 ms-1
s = ut + \(\frac{1}{2}\) at2
7 = 0 x 3 + \(\frac{1}{2}\) a x 32
7 = 0 + \(\frac{1}{2}\) a x 9
7 = \(\frac{9}{2}\)a
9a = 7 x 2
9a = 14
a \(\frac{14}{9}\) = 1.5ms-2
IV. Answer the following:
Question 1.
If a body is moving with uniform velocity in a given direction its acceleration Will be zero. Why?
Answer:
The acceleration will be zero in the above situation since there is no change in velocity in unit time.
Question 2.
Distinguish between speed and velocity
Answer:
Speed:
- Distance travelled by a body in unit time.
- It is a scalar quantity.
- Speed= Distance/Time.
- It has only magnitude.
Velocity:
- Velocity is the rate of displacement of the body in unit time.
- It is a vector quantity.
- Velocity = Displacement/time.
- It has both magnitude and direction.
Question 3.
Distinguish between distance travelled and displacement.
Answer:
Distance travelled:
- It is the length of the actual path travelled by a body from one position to another.
- It is a scalar quantity.
- It has only a numerical value.
Displacement:
- It is the shortest path between the initial and final positions of the moving body.
- It is a vector quantity.
- It has both numerical value and direction.
Question 4.
What are uniform and non-uniform speeds?
Answer:
a. if an object covers equal distances in ‘ equal intervals of time, it is said to be uniform speed.
b. If an object covers unequal distances in equal intervals of time, it is said to be non-uniform speed.
Question 5.
While mentioning acceleration the time is mentioned two times why?
Answer:
Velocity is the rate of displacement of a body in unit time
i.e. v = \(\frac{\text { Displacement }}{\text { Time }}\)
Acceleration is the change in velocity of a body in unit time.
i.e. = \(\frac{v-u}{t}\)
Therefore while mentioning acceleration time is mentioned two times.
V. Extended activity
Question 1.
Represent the following motion by a graph.
Velocity (ms-1) | 40 | 30 | 20 | 10 |
Time (s) | 1 | 2 | 3 | 4 |
1. Find the acceleration
2. Find the time taken when the velocity is 35 ms-1
Answer:
1. The graph shows the negative acceleration or retardation.
v = 30 ms-1
u = 30 ms-1
t = 1 sec
KSEEB Class 8 Science Describing Motion Additional Questions and Answers
Question 1.
Define motion.
Answer:
The change in position of a body with time when compared with that of another body is called motion.
Question 2.
Motion is relative – Substantiate this / statement with a suitable example.
Answer:
Imagine that you are sitting inside the train. When the train starts moving, you feel that the persons on the platform are moving backward and also feel that you and other persons inside the train are not moving. But for a person standing outside the train, the feeling will be that the train, you and other members inside the compartment are all moving.
Question 3.
When can we say that speed and velocity are synonyms.
Answer:
Speed and velocity will be equal for uniform motions along a straight line. Then they will be one and the same.
Question 4.
For an object moving at a uniform speed in a circular path, though the speed is uniform it is continuously accelerated. Why?
Answer:
For an object moving in a circular path, the direction is changing at every point even though the speed is the same.
Therefore velocity changes and hence it is continuously accelerated.
Question 5.
What is the distance-time graph? What are its uses?
Answer:
Motion can be represented by* time graphs. If it represents time on the x-axis and distance travelled on the y-axis, it is called the distance-time graph. From this graph,
- the distance at which the body is at present from its starting point can be found out the
- time taken to cover a definite distance can be determined.
- since it is uniform motion, the speed of the body can also be calculated.
Question 6.
What do you infer about the velocity when the velocity-time graph is parallel to the x-axis.
Answer:
Since it is parallel, the velocity will be uniform and it does not change. Therefore the acceleration also will be zero.
Question 7.
What is negative acceleration?
Answer:
Negative acceleration is the acceleration of a body when it is coming from motion to rest. i.e. the velocity of the body is decreasing. Negative acceleration is also called as retardation.
II. Fill in the blanks :
Question 1.
- Displacement is a _____________ quantity.
- Distance travelled is a _____________ quantity.
- The velocity of a body at rest _____________.
- The distance travelled in unit time is called _____________.
- S. I. unit of speed as _____________.
- S. I. unit of displacement is _____________.
- Speed and velocity are _____________ for uniform motions along a straight line.
- If the value of acceleration is positive, the velocity is _____________.
- If the value of acceleration is negative, then the velocity is _____________.
- The acceleration is _____________ when the velocity-time graph is parallel to the x-axis.
Answer:
- vector
- scalar
- zero
- speed
- metre per second (ms-1),
- metre
- equal
- increasing
- decreasing
- zero
III. Choose the best alternative :
Question 1.
The scientist who gave the theory of relativity
(a) Albert Einstein
(b) Isaac Newton
(c) Max Planck
(d) Johannes Kepler
Answer:
(a) Albert Einstein
Question 2.
The length of the path covered by a body is
(a) displacement
(b) distance
(c) speed
(d) velocity
Answer:
(b) distance
Question 3.
The displacement of a body after completing one revolution along the circular path is
(a) 2πr
(b) \(\frac{2 \pi r}{t}\)
(C) πr2
(d) Θ
Answer:
(d) Θ
Question 4.
Unit of acceleration is
(a) ms-1
(b) ms-2
(c) ms1
(d) ms2
Answer:
(b) ms-2
Question 5.
Formula to calculate acceleration of u,v and t are the initial velocity, final velocity and time respectively is
(a) a = (v + u)/t
(b) a = (u – v)/t
(c) a = (v – u)/t
(d) (v – u)/t
Answer:
(c) a = (v – u)/t
Question 6.
An insect moves 3 cm in a straight line towards the east and then moves 4 cm to the south. Its displacements are cm.
(a) 1
(b) 5
(c) 7
(d) 25
Answer:
(b) 5
Question 7.
Velocity has the same unit as that of
(a) acceleration
(b) speed
(c) retardation
(d) displacement
Answer:
(b) speed
Question 8.
A person goes from x to y with a velocity of 40 ms1 and comes back to A with a velocity of 60 ms1. The average velocity for the total journey is
(a) 0 ms-1
(b) 25 ms-1
(c) 50 ms-1
(d) 100 ms-1
Answer:
(c) 50 ms-1
Question 9.
A student goes to school which is 5 km away from home and returns home in the evening. The displacement of the entire trip has a magnitude of
(a) 10 km
(b) 5 km
(c) 15 km
(d) 0 km
Answer:
(d) 0 km
Question 10.
Speed equals
(a) distance/time
(b) time/distance
(c) distance x time
(d) time/acceleration
Answer:
(a) distance/time
Question 11.
Rate of change of velocity is
(a) speed
(b) displacement
(c) distance travelled
(d) Acceleration
Answer:
(d) Acceleration
Question 12.
Which one of the following is notan t equation of motion
(a) v = u + at
(b) v2 = u2 + 2as
(c) s = ut + 1/2 at2
(d) s = vt + 1/2 at2
Answer:
(d) s = vt + 1/2 at2
IV. Problems :
Question 1.
A train takes 3 hours to travel from Agra to Delhi with a uniform speed of a 70 km/h. Find the distance between the two cities.
Answer:
speed = 70 km/h time taken = 3 hr.
distance = sp x t = 70 x 3 = 210 km
Question 2.
A car starts for rest and acquires a velocity of 10 ms in 2 seconds. Find its acceleration.
Answer:
initial velocity of car ‘u’ = 0
final velocity of car ‘v’ = 10 ms-1
time = 2 sec
a = \(\frac{v-u}{t}=\frac{10-0}{2}\) = 5 ms-2
Question 3.
A body starts with an initial velocity of 10 ms-1 and acceleration 5 ms-1. Find the distance by it in 5 seconds.
Answer:
time taken = 5 seconds
initial velocity u = 10 ms-1
acceleration = 5 ms-2
distance travelled = ?
Question 4.
A motorcyclist starts from rest and attains a uniform acceleration of 9 km/h2. What will be the velocity at the end of 2 km distance?
Answer:
u = 0, a = 9 cm/h2
s = 2 km, v = ?
v2 = u2 +2as
v2 = a2 + 2(9) x 2
v2 = 36
v2 = \(\sqrt{36}\)
6km/h
Question 5.
A bus travels 43 km in the first hour, 40 km in the second hour and 46 km in the third hour of its journey. Calculate the average speed.
Answer:
average speed = \(\frac{\text { total distance }}{\text { time taken }}\)
= \(\frac{43+40+46}{3}\)
= \(\frac{120}{3}\)
= 4.3 km h-1
Question 6.
Place B is 450 km away from A. If a train starts from A at 6.15 am and travels continuously with an average speed of 50 km/hr. When does it reach B?
Answer:
Sp = 50 km/hr
d = 450 km
time taken = \(\frac{\mathrm{d}}{\mathrm{sp}}\) = \(\frac{450}{50}\) = 9 hrs.
The time at which the train reaches place
B = 6.15 + 9 = 15.15- 12 = 3.15 pm
Question 7.
A car which is moving with a velocity of 60 km/hr comes to rest in 5 sec. Calculate its acceleration in m/s2.
Answer:
u = 60 km/hr
\(=\frac{60 \times 1000}{3600}=\frac{100}{6}=\mathrm{ms}^{-1}\)
t = 5 sec
v = 0
\(a=\frac{v-u}{t}=\frac{0-\frac{100}{6}}{5}\)
\(=\frac{-100}{6 \times 5}=\frac{-100}{30}=-3 \frac{1}{3} \mathrm{ms}^{-2}\)
(retardation)