Are you looking for Motion Extra Questions and Answers, if yes? then you the right place. Here we are providing Motion Extra Questions and Answers Class 9 Science Chapter 8.
Motion Extra Questions and Answers Science Chapter 8
Very Short Answer Type questions:
1.) Is speed depend on direction of moving object?
Answer: No, speed is only depend on magnitude of path length and time.
2.) State the formula to calculate acceleration
Acceleration is calculated by a formula,
Acceleration = (final speed – initial speed)/time required to change in speed.
a = (v-u)/time
3.) State the SI unit of acceleration.
Answer: The SI unit of acceleration is m/s².
4.) What is the name of device which used to measure the speed of vehicle?
Answer: Speed of vehicle is measure by speedometer or odometer.
5.) Define motion.
Answer: When an object changes its position with surrounding and time is called motion.
6.) Define distance:
Answer: Total path covered by an object is called distance.
7.) Give the definition of displacement.
Answer: The shortest distance between initial and final position is called displacement.
8.) Give the formula to calculate average speed.
Answer: Average speed = (initial speed + final speed)/2
9.) Is acceleration scalar?
Answer: No, the acceleration is vector quantity.
10.) If the object moves circular path and complete one rotation then calculate it’s distance.
Answer: If the object moves circular path and complete one rotation then total distance in one rotation is equal to circumference of that circular track.
The total distance in one rotation = 2 ×π ×r.
11.) What is the value of displacement when an object completes 4 rotation?
Answer: The initial and final point of that object is same so the value of displacement becomes zero.
12.) What is the value of initial speed when a stone is dropped from certain height?
Answer: If we drop a stone from certain height then value of initial speed is zero.
13) Is motion of athletes who moves with constant speed along a circular track accelerated?
Answer: Yes, he moves with constant speed but changes direction continuouslyso it is an example of accelerated motion.
14.) What is the value of area under velocity time graph?
Answer: Area of time-velocity graph is displacement.
15.) Give the example of circular motion.
Answer: Example of circular motion
a) Motion of athletic on circular track.
b) Motion of tip of hand of clock.
16.) On what factor speed is depend?
Answer: Speed is depend on distance and time.
Short Answer Type questions:
1.) Is motion relative concept? Explain your answer with example.
Answer: Yes, motion is relative concept.
Example: When you sit in a moving train the co passengers are also in motion but according to you they are in rest. The trees outside the train appears in motion even the state of that tree are rest.
2.) State four examples of motion in straight line.
Answer: Example of motion in a straight line.
a) Motion of a plane before take off.
b) Motion of a car on straight road.
c) Motion of a swinging of pendulum.
d) Motion of a period of soldiers on straight track.
3.) If a man moves 5 km on straight road and moves back at initial position then calculate distance covered by that man.
Answer: The man covers 5 km in forward journey and backward also 5 km. And we know that distance must the total path length. Therefore,
total distance = 5 km + 5km
Distance = 10km.
4.) Convert 54 km/hr into m/s.
Answer: As we know that, 1 km = 1000 m and 1 hr =3600 seconds
54 km/hr = 54 × 1000/3600 m/s
54 km/hr = 54 × 5/18 m/s
54 km/hr = 3 × 5 m/s
54 km/hr = 15 m/s
5.) If the vehicle moves with 20 m/s and 40 m/s then calculate average speed of that vehicle.
Answer: Given, initial speed = u = 20 m/s and
Final speed = v = 40 m/s.
As we know that
Average speed = (u+v)/2
Average speed = (20 + 40)/2.
Average speed = (60/2)
Average speed= 30 m/s
6.) If a car covers first 200 m in 30 seconds and next 300 m in 20 seconds then calculate average speed of that car.
Answer: Given, first distance = d1 = 200 m and last distance d2 = 300 m.
Time required to cover first distance = t1 = 30 seconds and time required to cover last distance
t2 = 20 seconds.
As we know that,
average speed = (d1 + d2 )/(t1 + t2 )
Average speed = (200 + 300)/(30 + 20)
Average speed = 500/50
Average speed = 10 m/s.
The Average speed of that car is 10 m/s.
7.) Describe the concept of velocity.
Answer: When an object changes a position continuouslythen we say that the object is in motion. We describe this change with time. Velocity gives information about this.
8.) If a car accelerate from 10 m/s to 20 m/s in 2 seconds. Calculate acceleration of that car.
Answer: Given, initial speed =u = 10 m/s,
Final speed = v = 20 m/s, time = T = 2 s.
As we know,
Acceleration = (v-u)/t.
Acceleration = (20-10)/2
Acceleration =10/2
Acceleration =5 m/s².
The acceleration of that car is 5 m/s².
9.) If a stone is released from the top of tower and reach after 4 seconds. Calculate the height of tower.(g=10 m/s²)
Answer: Given, the stone is released so initial speed = u = 0 m/s,
Time = t = 4 seconds, g=10 m/s².
As we know, first equation of kinematics,
x = ut + ½ at²
x = 0 × t + ½ × 10 × 4²
x = 5 × 16
x = 80 m.
The height of that tower is 80 m.
10.) Derive first equation of kinematics.
Answer: We can derive first equation of kinematics from the definition of acceleration.
As we know,
Acceleration =(final speed – initial speed)/time.
a = ( v – u )/t.
at = v – u.
at + u = v
v = u + at.
This is the first equation of motion.
In case you are missed :- Previous Chapter Extra Questions
11.) Give four example of uniform circular motion.
Answer: Examples of uniform circular motion.
a) Motion of tip of hand of clock.
b) Motion earth around sun.
c) Motion of electron around nucleus.
d) Motion of satellites around planet.
12.) A particle moves from point ‘O’ to point ‘Q’ and returned at point ‘P’. As shown in figure. Calculate (a) Distance. (b) Displacement. (The unit of scale is in metre.)
Answer: (a) Distance is the actual path travelled by an object. There fore total distance is the path length from pint ‘O’ to point ‘Q’ and ‘Q’ to point ‘P’.
Distance = 10 + 4
Distance = 14 m
(b) Displacement is the smallest distance between initial position and final position.
Displacement = 6 m.
Answer: (a) Distance is the actual path travelled by an object. There fore total distance is the path length from pint ‘O’ to point ‘Q’ and ‘Q’ to point ‘P’.
Distance = 10 + 4
Distance = 14 m
(b) Displacement is the smallest distance between initial position and final position.
Displacement = 6 m.
13) If a stone throws upward with velocity 10 m/s then calculate maximum height gain by that stone. ( g= 10 m/s²)
Answer: Given, initial velocity = u = 0 m/s. At maximum height velocity becomes zero.
Final velocity = v = 0 m/s, the acceleration opposes the motion, Acceleration = -10 m/s².
As we know, third equation of kinematics,
v² = u² + 2 a x
0² = 10² + 2 ×(-10) × h
20h = 100
h = 100/20
h = 5 m.
The maximum height gain by that stone is 5 m.
14) Draw a velocity-time graph for accelerated motion of particle.
Answer: The velocity-time graph for accelerated motion is as,
15) The distance and magnitude of displacement is equal. State the condition to prove this statement.
Answer: If the particle moves in straight line without changes the direction. In this case the distance and magnitude of displacement is equal.
Long Answer Type Questions:
1) (a) A particle moves along a road. The speedometer measures the reading are as-
Sr No | Time (s) | Velocity |
1 | 0 | 0 |
2 | 2 | 10 |
3 | 4 | 20 |
4 | 6 | 30 |
5 | 8 | 40 |
6 | 10 | 50 |
7 | 12 | 50 |
8 | 14 | 70 |
9 | 16 | 80 |
Plot a velocity-time graph.
(b) Is it uniform motion. Give the reason.
Answer: Reading of velocity and time are as above. We plot a graph from the above data.
(b) The motion performed by the particle is uniform motion because the graph is linear.
2) The graph of motion of the car is as shown in figure.
Calculate,
a) Distance travelled by car in 28 seconds.
b) Acceleration of car in first 13 seconds.
Answer: (a) As we know that, Area under time speed graph gives the value of distance.
Distance = area of triangle
But area of triangle = ½ ×base × height
Distance = ½ (28 × 16)
Distance = 28 × 8
Distance = 224 m.
Car covers 224 m in 28 seconds.
(b) From above graph,
Initial speed = u = 0 m/s,
Final speed = v = 12 m/s, time = 12 seconds.
As the definition of acceleration,
Acceleration = (v – u)/t
Acceleration = (12-0)/12
Acceleration = 1 m/s².
The acceleration of the car is 1m/s².
3) (a) Give the second equation of motion and explain each term in it.
(b) A car moves with constant speed of 10 m/s. The driver accelerate the car at the rate of 2 m/s² for 5 seconds. Calculate total distance travelled by that car.
Answer:(a) The second equation of motion gives the relationship between distance, initial speed, Acceleration and time. The second equation of kinematics is as,
x = ut + ½ a t²,
where,
x = displacement,
u – initial speed,
t – Time,
a – acceleration,
(b) Given, initial speed = u = 10 m/s, time = 5 seconds, Acceleration = a = 2 m/s².
As we know second equation of motion,
x = ut + ½ a t²
x = 10 × 5+ ½ × 2 × 5²
x = 50 + 25
x = 75 m.
The car moves 75 m in 5 seconds.
4) (a) Give the third equation of motion and explain each term in it.
(b) A car is moving with speed 20 m/s. A pole is 100 m away from the car so driverdecelerate the car and stop it before hit the pole. Calculate deceleration of the car.
Answer: (a) The third equation of motion gives the relationship between final speed, initial speed, distance and acceleration. The third equation of motion is as,
v² = u² + 2 ax
Where,
v –final speed,
u – initial speed,
a – acceleration,
x – distance.
(b) Given, initial speed = u = 20 m/s, Finally car stop so final speed = 0 m/s,
Distance = 100 m, deceleration = -a.
As we know third equation of kinematics,
v² = u² + 2 ax
0² = 20² + 2× (-a) × 100
200a = 400
a = 400/200
a = 2 m/s².
The driver must decelerate the car at rate of 2 m/s².
5) (a) Explain uniform circular motion.
(b) If a body moves along circular track of radius 42 m and complete 4 rotation in 60 seconds then calculate speed and velocity.
Answer: (a) When an object moves along a circular track with constant speed then such motion is called as uniform circular motion. The particle perform circular motion with constant speed not the constant velocity. At every point the speed remains constant but direction changes continuously so the value of velocity changes continuously. The uniform circular motion is accelerated motion.
(b) Given, radius = 42 m, no of rotation = 4, time = 60 seconds.
As we know that the speed in circular motion
=( 2 × π ×r)/t
The car completes 4 rotation so total distance becomes 4 × circumference of circle.
Speed = 4× (2 × 22/7 × 42)/60
Speed = (8 × 22 × 6)/60
Speed = 17.6 m/s.
The speed of that car is 17.6 m/s.
The car completes 4 rotation and reach at initial point therefore displacement becomes zero and thus value of velocity also becomes zero.
In case you are missed :- Next Chapter Extra Questions