NCERT CLASS 9 SCIENCE CHAPTER 7 MOTION COMPLETE SOLUTION - PARAM HIMALAYA
Class 9 Science Chapter 7 motion Page number 74 Question 1 to 3 Solution with video Explanation :
Question 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Solution :
Yes, An object instead of moving through a distance can have zero displacement.
Example: If an object travels from point A and reaches to the same point A, then its displacement is zero.
Total Distance = 4 + 4 = 8 m
Distance = Final Position - Initial Position
= 0-0 = 0 m
Question 2. 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?
Solution :
Given Side of square field = 10m
So, Perimeter of a square = 4 × side = 4×10 m = 40 m
Farmer takes 40 s to move along the boundary.
Displacement after 2 minutes 20 s = 2 × 60 s + 20 s = 140 seconds
Since in 40 s farmer moves 40 m
Therefore, in 1s the distance covered by farmer = 40 / 40 m = 1m
Therefore, in 140s distance covered by farmer = 1×140 m = 140 m.
Now, number of rotation to cover 140 along the boundary= Total Distance / Perimeter = 140 m / 40 m = 3.5 round
Thus, after 3.5 round farmer will at point C of the field.
Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14 m north east from initial position.
Question 3. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object.
Solution :
Both (a) as well as (b) are false with respect to concept of displacement.
Example 7.1 An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the object?
Solution:
Total distance travelled by the object = 16 m + 16 m = 32 m
Total time taken = 4 s + 2 s = 6 s
Average speed = Total distance travelled / Total time taken
= 32 m / 6 s = 5.33 m / s
Therefore, the average speed of the object is 5.33 m / s
Class 9 Science Chapter 7 motion Page number 76 Question 1 to 3 Solution with video Explanation :
Question 1. Distinguish between speed and velocity.
Solution :
Question 2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Solution :
If distance travelled by an object is equal to its displacement then the magnitude of average velocity of an object will be equal to its average speed.
Question 3. What does the odometer of an automobile measure?
Solution :
The odometer of an automobile measures the distance covered by that automobile.
Question 4. What does the path of an object look like when it is in uniform motion?
Solution :
Graphically the path of an object will be linear i.e. look like a straight line when it is in uniform motion.
Question 5. 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 × 108m/s.
Solution:
Given,
Speed of the signal = 3×108m/s
Total time taken by the signal to reach earth
Time = 5 min = 5× 60 s = 300 s
We know distance travelled can be given as
Distance = Speed×Time
Putting all the values and solving
Distance = 3×108 m/s ×300 s = 9×1010 m
Hence the distance of the spaceship from the earth is 9×1010 m
Example 7.2 The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 h, calculate the average speed of the car in km/h and m/s.
Solution:
Distance covered by the car,
s = 2400 km – 2000 km = 400 km
Time elapsed, t = 8 h
Average speed of the car is
Vav = s/t = 400 km / 8 h = 50 km/h
= 50×5/18 = 250/18 m/s = 13.9 m/s
The average speed of the car is 50 km/h or 13.9 m/s.
Example 7.3 Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.
Solution:
Total distance covered by Usha in 1 min is 180 m.
Displacement of Usha in 1 min = 0 m
Average Speed = Total distance covered/Total time taken
= 180m/1min = 180 m/160 s
= 3 m / s
Average Speed = Displacement / Total time taken
= 0 m / 60 s
= 0 m/s
The average speed of Usha is 3 m/s
and her average velocity is 0 m /s.
Example 7.4 Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m s-1 in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m s-1 in the next 5 s. Calculate the acceleration of the bicycle in both the cases.
Solution :
In the first case:
initial velocity, u = 0 , final velocity, v = 6 m/s , time, t = 30 s
we know that
a = (v – u)/t
Substituting the given values of u,v and t in the above equation, we get
a= (6m/s – 0m/s) / 30 s = 0.2 m /s2
In the second case:
initial velocity, u = 6 m s–1;
final velocity, v = 4 m s–1;
time, t = 5 s.
a = (4m/s – 6m/s) / 5s = –0.4 m/s2 .
The acceleration of the bicycle in the
first case is 0.2 m/s2 and in the second
case, it is –0.4 m/s2.
Class 9 Science Chapter 7 motion Page number 77 Question 1 to 3 Solution with video Explanation :
Question 1. When will you say a body is in (i) uniform acceleration? (ii) nonuniform acceleration?
Solution :
(i) Uniform acceleration :
When a body is changing its velocity at a constant rate means equal change in velocity in an equal interval of time, then the body is said to be moving with uniform acceleration.
Example: the free-falling of an object.
(ii) Non-uniform acceleration :
When a body is changing its velocity at different rates, then the body is said to be moving with non-uniform acceleration.
Example: The bus is leaving from the stop.
Question 2. A bus decreases its speed from 80 km / h to 60 km / h in 5 s. Find the acceleration of the bus.
Solution:
Given: Initial velocity (U) = 80 km/hr = 80 × 5/18 = 400/18 = 22.22 m/sec
Final velocity (V) = 60 km/hr = 60 × 5/18 = 300/18 = 16.67 m/sec
Acceleration a = (V - U) / t
= (16.67 - 22.22)/5
= -5.55/5
= -1.11 m/s2
Negative sign indicates that the velocity is decreasing.
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:
Given :
Initial speed of the train, u = 0 km (Because the train starts from the rest).
Final speed of the train, v = 40 km/h = (5/18) × 40 = 11.11 m/s.
Time taken to attain the speed = 10 minutes
= 10 × 60
= 600 sec
We know that
acceleration a = (v−u)/ t
a = (11.11−0)/ 600 = 0.0185 m/s²
Class 9 Science Chapter 7 motion Page number 81 Question 1 to 3 Solution with video Explanation :
Question 1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Solution:
The Distance-Time Graph for uniform motion is a Straight line.
The Distance - Time Graph for Non-uniform motion is a curved line.
Question 2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Solution:
If the distance-time graph is a straight line parallel to time axis , that means the body is not changing its position with time , the body is at rest.
Question 3. What can you say about the motion of an object if its speed time graph is a straight line parallel to the time axis?
Solution:
When the slope of a speed time graph is a straight line parallel to the time axis, the object is moving with uniform speed.
Example 7.5 A train starting from rest attains a velocity of 72 km h–1 in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity.
Solution:
Given :
v = 72 km/h ( Final velocity ) = 72 × (5/18) = 20 m/s
u = 0 ( initial velocity )
t = 5 min ( time ) = 5×60 = 300 sec
(i) Acceleration a = ?
v = u + at
20 = 0 + a (300)
a = 20/300 = 1/15 m/S2.
a = 1/15 = 0.066 m/S2
(ii) Distance S = ?
S = ut + 1/2 at2
S= 0 + 1/2 (1/15) (300)2
S = 90000/30
S = 3000m
Example 7.6 A car accelerates uniformly from 18 km h–1 to 36 km h –1 in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.
Solution:
Given
Initial velocity, u = 18 km/hr = 5 m/s
Final velocity, v = 36 km/hr = 10 m/s
Time , t = 5 s
(i ) Acceleration, a = ?
Using the first equation of motion :
v = u + at
10 = 5 + a (5)
a = 1m/S2
(ii) Distance , S = ?
S = ut + 1/2 at2
= 5×5 + 1/2 (1) 5
S = 37.5 m
Example 7.7 The brakes applied to a car produce an acceleration of 6m/s2 in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.
Solution:
Given :
Acceleration a = -6m/s2
Time t = 2 s
Final velocity v = 0 m/s
Initial velocity u = ?
Distance s = ?
From the first equation of motion,
v = u + at
0 = u + (-6) 2
u = 12 m/s
From the third equation of motion ,
v2 - u2 = 2as
02 - 122 = 2(-6)s
-144 = -12 s
Or
-12 s = -144
s = 144/12
s = 12 m
Class 9 Science Chapter 7 motion Page number 82-83 Question 1 to 5 Solution with video Explanation :
Question 1. A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Solution:
Given :
Initial velocity (u) = 0 m/s
Acceleration (a) = .1 m/s2
Time (t) = 2 min = 120 second
(i) the speed acquired (v) = ?
v = u + at
v = 0 + .1 × 120
v = 12 m/s
(ii) The distance travelled (s) = ?
s = ut + 1/2 at2
s = 0×120 + 1/2 × .1 × (120)2
s = 1/2 × .1 × 14400
s= 720 m
Question 2. A train is travelling at a speed of 90 km h–1. Brakes are applied so as to produce a uniform acceleration of – 0.5 m s-2. Find how far the train will go before it is brought to rest.
Solution:
Given :
Initial velocity u = 90 km/h = 90 × (5/18)
= 450/18 = 25 m/s
Final velocity , v = 0 m/s
Acceleration, a= - 0.5 m/ s2
Distance travelled = ?
Using v2 - u2 = 2as
s = ( v2 - u2 ) / 2a
s = ( 02 - 252 ) / 2 (-.5) = 625 m
s = 625 m
Question 3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2. What will be its velocity 3 s after the start?
Solution:
Given :
Initial velocity, u = 0 m/s
Acceleration (a) = 2 cm/s2 = 0.02 m/s2
Time (t) = 3s
Final velocity, v =?
We know that
v = u + at
v = 0 + 2 ×3
v = 6
v = 6 sec
Question 4. A racing car has a uniform acceleration of 4 m /s2. What distance will it cover in 10 s after start?
Solution:
Given a = 4 m/s2
t = 10s
u = 0
s = ut+½ at2
s = 0×10 + ½ 4 (10)2
s = (4×100) /2
s = 400/2
s = 200m
Question 5. A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Solution :
Given :
u = 5 m/s
a = - 10 m/ s2
v = 0 m/s2
H = ?
t = ?
We know
v = u + at
0 = 5 + (-10)t
-5 = -10t
t = 5/10
t = .5 sec
Now
v2 = u2 + 2as
0 = 52 + 2(-10)s
0 = 25 -20s
25= 20s
s = 25/20
s = 1.25 m
Exercise Question :
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:
The diameter of the circular track = 200 m.
2r = 200 m.
r = 200 m/2 = 100 m.
In 40 sec the athlete complete one round.
So, in 2 mins and 20 secs, that is, 140 sec the athlete will complete = 140 / 40 = 3.5 (three and a half) rounds.
One round is considered as the circumference of the circular track.
The distance covered in 140 sec = 2πr×3.5 = 2×3.14×100×3.5 = 2200 m.
For each complete round the displacement is zero. Therefore for 3 complete rounds, the displacement will be zero.
At the end of his motion, the athlete will be in the diametrically opposite position.
That is, displacement = diameter = 200 m.
Hence, the distance covered is 2200 m and the displacement is 200 m.
Question 2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 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:
From point A to B :
Distance covered=300m
Displacement=300m
Time taken=2 min 30 sec=(2×60)+30=150 sec
Average speed= Distance covered / Time Taken = 300 /150 = 2 m/sec
Average velocity= Displacement / Time Taken = 2 m/sec
From point A to C :
Distance covered=300+100=400m
Displacement=300−100=200m
Time taken=3 min 30 sec=(3×60)+30=210 sec
Average speed= Distance covered / Time Taken = 400 / 210 = 1.90 m/sec
Average velocity= Displacement / Time Taken = 200 / 210 = .95 m/sec
Question 3. Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed for Abdul’s trip?
Solution :
Case I: While driving to school
Average speed of Abdul’s trip = 20 km/h
Total distance = d
Let total time taken = t1
Speed = Distance / time
20 = d / t1
t1 = d / 20 - (i)
Case II: While returning from school
Total distance = d
Speed = 30 km/h
Now, total time taken = t2
Speed = distance / time
30 = d / t2
t2 = d / 30 - (ii)
Total distance covered in the trip = d + d = 2d
Total time taken, t = Time taken to go to school + Time taken to return to school
= t1 + t2
Total time taken = d / 20 + d / 30
= 3d + 2d / 60
= 5d / 60
= d / 12
Average Speed = 2d / (t1 + t2)
From equations (i) and (ii),
Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed for Abdul’s trip?
Average Speed = 120/5 = 24 km/h
Hence, the average speed for Abdul’s trip is 24 km/h.
Question 4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s2 for 8.0 s. How far does the boat travel during this time?
Solution :
Given acceleration a = 3 m/s2
Time t = 8s
Initial Velocity u = 0
Find Distance (S) =?
So, using the equation we get
S = ut+½ at2
S = 0 +½ (3×82)
S = 192/2
S = 96m
Question 5. A driver of a car traveling at 52 km/h applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km/h in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars traveled farther after the brakes were applied?
Solution :
Case A:
Initial speed of the car,
u1 = 52 km/h = 52 x (5 / 18) = 14.44 m/s
Time taken, t1 = 5 s
Final speed = 0 m/s
Case B:
Initial speed of the car,
u2 = 3 km/h = 3 x (5 / 18)
= 0.833 m/s ≅ 0.83 m/s
Time taken, t2 = 10 s
Final speed = 0 m/s
On plotting Graph
Distance covered by each car is equal to the area under the speed−time graph.
Distance covered in case A,
S = 1/2 ×OP × OR = 1/2 × 14.4 × 5 = 36 m
Distance covered in case B,
S2 = 1/2 × OS × OQ = 1/2 × .83 × 10 = 4.15 m
Then, the cart 1 travelling with a speed of 52 km/h travels farther after brakes were applied.
Question 6. Fig 7.10 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions.
(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?
(a)Speed : Slope (angle) of graph for B is maximum
(b) No , no intersection point of A,B and C.
(c) When B passes A, C is at 8 Km
Initially (at t=0) C was at $\frac{16}{7}$
Distance travelled by C = $8-\frac{16}{7}=\frac{40}{7}$
(d) When B passes C , B is at $\frac{36}{7}km$
initially (at t=0) B was at 0 Km
Distance travelled by B = $\frac{36}{7}$
Question 7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground?
Solution :
Given
u = 0 m/s , s = 20 m , a = 10 m/s2 ,
v= ? , t =? ,
On using v2 = u2 +2as
We have , v2 = 02 + 2×10×20
v2 = 400
v = 20 m/s
On using
v = u+at
20 = 0 + 10×t
t = 20/10
t = 2 sec
Question 8. The speed-time graph for a car is shown is Fig. 7.11.
(a) Find how far does the car travel in the first 4 seconds.
Shade the area on the graph that represents the distance
travelled by the car during the period.
(b) Which part of the graph represents uniform motion of
the car?
Solution:
(a) The area enclosed by speed -time graph and the time axis is equal to the magnitude of the distance covered.
Distance (approx) = $\frac{1}{2}(4)(6)= 12m$
(b) The car moves under the uniform motion after the instant (t=6 sec) because after this instant, the speed of the car becomes constant.
Question 9. State which of the following situations are possible and give an example for each of these:
(a) an object with a constant acceleration but with zero velocity
(b) an object moving with an acceleration but with uniform speed.
(c) an object moving in a certain direction with an acceleration in the perpendicular direction.
Solution :
(a) Possible, When a ball is thrown up, at the highest point, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s2.
(b) Possible, When a car is moving in a circular track with constant speed, it is accelerating due to change in direction of motion.
(c) Possible, A car is moving in a circular track, its acceleration is perpendicular to the direction of motion.
Question 10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Solution :
Given that :
Radius of circular orbit r = 42250 km
Time period of satellite T = 24 hours
Distance traveled by satellite to complete 1 revolution = Circumference of a circle
S = 2πr
S = 2 x 3.14 x 42250 = 265464.58 km
When the distance D and time T are given, speed v can be calculated using
v = D/T
v = 265464.58/(24 x 60 x 60)
v = 3.07 km/s
Therefore, the speed is 3.07 km/s.