ncert solutions for class 9 science
ncert solutions for class 9 science chapter 8 exercise, motion chapter class 9 science ncert solution all question in full detailed explain in easy language. ncert motion chapter page no. 100, 103, 107, 102, 109, 110, and all questions.
ncert solutions for class 9 science chapter 8 exercise
Chapter Motion NCERT Book Page 100
Q. An object has moved through a distance. Can it has Zero displacement? If yes, support your answer with an example.
Answer:- yes it may be zero displacement because if an object moves some distance and come back to the initial position from where he start then the final and initial position will be same then the displacement will be zero. for example If we run in a circular path and after completing one round, come back to starting position then the displacement will be zero. another example if we go in a straight line and come back to follow the same path in initial position then the displacement is zero. note that displacement will be always zero whenever the initial and final position of the object will be same.
Q. 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 from his initial position?
Answer:- firstly we convert 2 minutes 40 seconds into seconds
Total time = 2 ✖ 60 seconds + 20 seconds
= 120 seconds + 20 seconds
= 140 seconds
Now, In 40 s, number of rounds made = 1
so, In 140 s, number of rounds made = 1/40 ✖ 140
= 3.5 rounds
the farmer will make three complete round and also half rounds ( 3.5 rounds ) of square field. If the farmer start from point ‘P’ as shown in fig. then after completing three round he will reach at the same point ‘P’. but he has to complete remaining half round so he complete remaining half round and reach at point ‘R’. Now the total displacement of the farmer will be ‘PR’.
Now, in triangle PQR, PR is the hypotenuse.
( PR )2 = (PQ)2 + (QR)2
( PR )2 = (10)2 + (10)2
( PR )2 = 100 + 100
( PR )2 = 200
PR = √200
PR = 14.143
Thus, the magnitude of displacement of the farmer at the end of 2 minutes and 20 seconds will be 14.143 meters
.
Q. Which of the following is true for displacement?
- It cannot be zero.
- Its magnitude is greater than the distance travelled by the object.
Answer:- (a) yes displacement can be zero. when any body go in a straight line or in circular motion then the displacement will be zero. it means that whenever the initial and final position of the body will same then the displacement will be zero.
Answer:- (b) no its magnitude can never be greater than the distance travelled by the body.
ncert solution for class 9 chapter 8 page 102
Q. Distinguish between speed and velocity.
Speed:-
- speed of a body is distance travelled by it per unit time.
- In speed direction of motion is not specified.
- so, speed has only magnitude not direction.
Velocity:-
- Velocity of a body is displacement travelled by it per unit time.
- In velocity direction is specified.
- so, velocity has both magnitude and direction.
Q. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Answer:- the magnitude of average velocity of an object equal to its average speed when the body go in a straight line path.
Q. What does the odometer of an automobile measure?
Answer:- odometer of an automobile measure the distance travelled by the vehicle.
Q. What does the path of an object look like when it is in uniform motion?
Ans:- An object is said to be in uniform motion when it cover equal distance in equal interval of time no matter how small these time interval may be. this means that the speed of an object is constant but the direction may be change. it may be circular, straight, zig-zag path, curved etc.
Q. 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.
Ans:- we know that,
Speed = distance / time
3 ✖️ 10^8 m/s = distance / 5 minutes
3 ✖️ 10^8 m/s = distance / 5 ✖️ 60 seconds
3 ✖️ 10^8 m/s = distance / 300
distance = 3 ✖️ 10^8 m/s ✖️ 300
distance = 9 ✖️ 10^10 m.
Thus, the distance travelled by the spaceship is 9 ✖️ 10^10 m.
ncert solutions for class 9 science chapter 8 page 103
Q. When will you say a body is in :
- uniform acceleration?
- Non-uniform acceleration?
Ans:- (a) A body is said to be in uniform acceleration if its velocity change uniformly with respect to time.
Ans:- (b) A body is said to be in non-uniform acceleration if its velocity change by unequal amount in equal amount of time or velocity change by equal amount in unequal interval of time then the body is in non-uniform acceleration.
Q. A bus decreases its speed from 80 km/h to 60 km/h in 5 sec. Find the acceleration of the bus?
Ans:- we know that,
acceleration = change in velocity / time
acceleration = u – v /t
acceleration = 80 – 60/ 5 seconds
acceleration = 16.66 – 22.22 / 5
a = – 1.11 m/s
Q. A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. find the acceleration.
Ans:- Initial speed of train, u = 0
Final velocity, v = 40 km/h
= 40 ✖ 1000 m/ 60 ✖ 60 s
= 11.11 m/s
Time taken, t = 10 minutes
= 10 ✖ 60 seconds
= 600 s
Acceleration, a = v – u/t
a = 11.11 – 0/ 600
a= 11.11/ 600
a = 0.0185 m/s2
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