NCERT Motion Class 9 Solution

ncert class 9 motion solution


ncert class 9 motion solution
Here are few problem solution for NCERT Motion class 9 solution. each solution is detailed enough to clear all the doubt

I  have given them some short and some long type questions

Short question:

Question   1:

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

Solution:

Uniform Acceleration: A body is said to have uniform acceleration if it travels in a straight path in such a way that its velocity changes at a uniform rate, i.e., the velocity of a body increases or decreases by equal amounts in an equal interval of time.

Non- uniform acceleration: A body is said to have non-uniform acceleration if it travels in a straight path in such a way that its velocity changes at a non-uniform rate, i.e., the velocity of a body increases or decreases in unequal amounts in an equal interval of time.

Question 2:

A bus decreases its speed from 80 km h?1 to 60 km h?1 in 5 s. Find the acceleration of the bus.

Solution:

u=80km/h=80 X5/18 m/s=200/9 m/s

v= 60 km/h=60X 5/18 m/s= 150/9 m/s

t=5 sec

Now we know that

V=u+at

So

a= (v-u)/t

Substituting the values

a= -10/9 m/s2

The minus sign here signifies that it is deceleration or retardation i.e. speed decreases

Question 3:

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

Solution :

Initial velocity of the train, u = 0 (since the train is initially at rest)

Final velocity of the train, v = 40 km/h = 40X5/18 m/s=100/9 m/s

Time taken, t = 10 min = 10 × 60 = 600 s

Now we know that

V=u+at

So

a= (v-u)/t

Substituting the values

a= 1/54 m/s2

Long answer Type questions

Question 1:

An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Solution:

We know that

Distance in physics, is the length of the path (the line or curve) described by an object moving through space. Distance is independent of direction

When a body moves from one position to another the shortest distance between the initial and final position of the body along with its direction is known as displacement

So Diameter of a circular track, d = 200 m

Circumference = ?d = ? (200) = 200? m

In 40 s, the given athlete covers a distance of 200? m.

In 1 s, the given athlete covers a distance = 200 ?/40=5 ?

The athlete runs for 2 minutes 20 s  which is equivalent to  140 s

Total distance covered in 140 s = 5 ?X140=2200m

Now calculating displacement is little bit tricky. We know athlete come to  its initial point in 1 round. So displacement would be zero. The athlete covers one round of the circular track in 40 s. This means that after every 40 s, the athlete comes back to his original position. Hence, in 140 s he had completed 3 rounds of the circular track and is taking the fourth round.

He takes 3 rounds in 40 × 3 = 120 s. Thus, after 120 s his displacement is zero.

Then, the net displacement of the athlete is in 20 s only. Now 40 sec ,he makes full round, so 20 sec he will be doing half round and reaching the opposite end of its initial position. Since displacement is equal to the shortest distance between the initial and final position of the athlete, displacement of the athlete will be equal to the diameter of the circular track.

Displacement of the athlete = 200 m

 

Question 2:

Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

Solution:

First of all we know that

Average velocity=Total displacement /Total  time

Average speed=Total distance  /total time



a)

Distance covered by Joseph while jogging from A to B = 300 m

Time taken to cover that distance = 2 min 50 seconds = 170 s

Total distance covered = 300 m

Total time taken = 170 s

So Average speed=300/170=1.765m/s

 

Displacement covered by Joseph while jogging from A to B = 300 m

Time taken to cover that distance = 2 min 50 seconds = 170 s

Total displacement covered = 300 m

Total time taken = 170 s

So Average velocity=300/170=1.765m/s

 

b)

Distance covered by Joseph while jogging from A to C = 300+100=400 m

Time taken to cover that distance = 2 min 50 seconds +1 min= 230 s

Total distance covered = 400 m

Total time taken = 230 s

So Average speed=400/230=1.739m/s

 

Displacement covered by Joseph while jogging from A to C = 300-100=200 m as he moves 100 min backward

Time taken to cover that distance = 2 min 50 seconds +1 min= 230 s

Total displacement covered = 200 m

Total time taken = 230 s

So Average speed=200/230=.87m/s

 

Question 3: Abdul, while driving to school, computes the average speed for his trip to be 20 km h?1. On his return trip along the same route, there is less traffic and the average speed is 40 km h?1. What is the average speed for Abdul’s trip?

Solution :

Let distance travelled by Abdul while driving to school =d

Let a and b are time taken during forward and backward journey

While driving to school

Average speed of Abdul’s trip = 20 km/h

So  20=d/a

Or a=d/20

While driving back from school

Average speed of Abdul’s trip = 40 km/h

So  40=d/b

Or

b=d/40

Average speed for the whole journey is given by

Average speed=  2d/(a+b) =2d/(20/d+ 40/d)=26.67 m/s

Question 4:

A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s?2 for 8.0 s. How far does the boat travel during this time?

Solution:

Initial velocity, u = 0 (since the motor boat is initially at rest)

Acceleration of the motorboat, a = 3 m/s2

Time taken, t = 8 s

According to the second equation of motion:

S = ut + (1/2)at2

Distance covered by the motorboat, s

S=0+(1/2) 3 *8*8= 96m

Hence, the boat travels a distance of 96 m.

 

 

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