Class 9 NCERT Solutions Physics Chapter 7 - Motion
Motion Exercise 74
Solution 1
Yes. An object that has moved through a distance can have zero displacement. Displacement is the shortest distance between the initial and the final position of an object. An object which has covered a distance can have zero displacement if it comes back to its starting point i.e., the initial position.
Consider the following situation. A man is walking along the boundary of a square park of side 20 m (as shown in the following figure). He starts walking from point A and after moving along all the sides of the park (AB, BC, CD, DA), he again comes back to the same point i.e., A.
In this case, the total distance covered by the man is 20 m + 20 m + 20 m + 20 m = 80 m. However, his displacement is zero because the shortest distance between his initial and final position is zero.
Solution 2
In 2 min and 20 s (140 s), he will cover a distance =
![begin mathsize 14px style 40 over 40 cross times 140 end style](https://images.topperlearning.com/topper/tinymce/cache/5a26ccb3f8f815e31bc2db9e1f5aef1c.png)
Therefore, the farmer completes
![begin mathsize 14px style 140 over 40 end style](https://images.topperlearning.com/topper/tinymce/cache/63622e8ecc8b564151c0efec709d183b.png)
That means, after 2 min 20 s, the farmer will be at the opposite end of the starting point.
Now, there can be two extreme cases.
Case I: Starting point is a corner point of the field.
In this case, the farmer will be at the diagonally opposite corner of the field after 2 min 20 s.
Therefore, the displacement will be equal to the diagonal of the field.
Hence, the displacement will be
![begin mathsize 14px style square root of left parenthesis 10 right parenthesis squared space plus space left parenthesis 10 right parenthesis squared end root end style](https://images.topperlearning.com/topper/tinymce/cache/32c145d47ac94774f24057b7cb08afd9.png)
Case II: Starting point is the middle point of any side of the field.
In this case the farmer will be at the middle point of the opposite side of the field after 2 min 20 s.
Therefore, the displacement will be equal to the side of the field, i.e., 10 m.
For any other starting point, the displacement will be between 14.1 m and 10 m.
Concept Insight: - Be careful about considering the cases, as the displacement in both the cases is different.
Solution 3
Displacement can become zero when the initial and final positions of the object are the same.
Concept Insight - Displacement is always less than or equal to the distance covered.
Motion Exercise 76
Solution 1
Speed |
Velocity |
Speed is the distance travelled by an object per unit time. It does not have any direction. |
Velocity is the displacement of an object per unit time. It has a unique direction. |
Speed is a scalar quantity. |
Velocity is a vector quantity. |
The speed of an object can never be negative. At the most, it can become zero. This is because distance travelled can never be negative. |
The velocity of an object can be negative, positive, or equal to zero. This is because displacement can take any of these three values. |
Solution 2
If the total distance covered by an object is the same as its displacement, then its average speed would be equal to its average velocity, i.e. when the object moves along a straight line path.
Concept Insight - Distance and displacement may or may not be equal to each other.
Solution 3
Solution 4
Solution 5
= 5 min = 5 × 60 = 300 s
Concept Insight - Convert all the quantities in the same units and then proceed to calculations.
Motion Exercise 77
Solution 1
(i) 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 the body increases or decreases by equal amounts in an equal intervals of time. The motion of a freely falling body is an example of uniform acceleration.
(ii) A body is said to have non-uniform acceleration if its velocity changes at a non-uniform rate, i.e., the velocity of the body increases or decreases by unequal amounts in an equal intervals of time. The motion of a car on a crowded city road is an example of non-uniform acceleration.
Solution 2
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/828_8154da0fcda7d4a11021f25351ffaf85.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/828_7d42e6fca65ce092c284f2bcffd50172.png)
Solution 3
![begin mathsize 14px style 40 cross times 5 over 18 equals 11.11 space straight m divided by straight s end style](https://images.topperlearning.com/topper/tinymce/cache/eceaed25460c729f860f744b7ec2f05e.png)
Time taken, t = 10 min =
![begin mathsize 14px style Acceleration comma space straight a equals fraction numerator straight v minus straight u over denominator straight t end fraction equals fraction numerator 11.11 minus 0 over denominator 600 end fraction equals 0.0185 space straight m divided by straight s squared end style](https://images.topperlearning.com/topper/tinymce/cache/f851ffaa53d6be5337f17adf35545464.png)
Motion Exercise 81
Solution 1
Solution 2
Solution 3
Solution 4
Let the velocity of the body at time (t) be v.
Area of the shaded region = length × breath
Length = t
Breath = v
Area = vt = velocity × time ...(i)
We know,
![begin mathsize 12px style Velocity equals Distance over Time end style](https://images.topperlearning.com/topper/tinymce/cache/83f6430dfeb9f7358380817730c89f45.png)
Hence, the area occupied below the velocity-time graph measures the distance covered by the body.
Motion Exercise 82
Solution 1
(12)2 - (0)2 = 2(0.1) s
So, speed acquired by the bus is 12 m/s.
Distance travelled by the bus is 720 m.
Solution 2
Solution 3
Acceleration, a = 2 cm s-2 = 0.02 m/s2
Time, t = 3s
According to the first equation of motion:
v = u + at
Concept Insight - Choose the equation of motion wisely out of the three, to minimize the number of steps in calculations.
where, v is the velocity of the trolley after 3s from start
v = 0 + 0.02 × 3 = 0.06 m/s
Hence, the velocity of the trolley after 3s from start is 0.06 m/s.
Motion Exercise 83
Solution 4
Acceleration, a = 4 m/s2
Time taken, t = 10 s
According to the second equation of motion:
where, s is the distance covered by the racing car
Solution 5
Final velocity, v = 0 (since the stone comes to rest when it reaches its maximum height)
Acceleration of the stone, a = acceleration due to gravity, g = 10 m/s2 (in downward direction)
There will be a change in the sign of acceleration because the stone is being thrown upwards.
Concept Insight - Change in sign of the acceleration due to change in direction is crucial as it'll change the results.
Acceleration, a = -10 m/s2
Let s be the maximum height attained by the stone in time t.
According to the first equation of motion:
v = u + at
0 = 5 + (-10) t
v 2 = u 2 + 2 as
(0) 2 = (5) 2 + 2(-10) s
Motion Exercise 85
Solution 1
![begin mathsize 12px style straight r equals straight d over 2 equals 100 space straight m end style](https://images.topperlearning.com/topper/tinymce/cache/a6c4fcc6d06e1a27f64aa7fc70d827c9.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/835_ee5cbe4f476b3dbda8d8f2dce0411466.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/835_ee5cbe4f476b3dbda8d8f2dce0411466.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/835_ee5cbe4f476b3dbda8d8f2dce0411466.png)
Concept Insight - Circumference of a circle is given by (2 ×
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/835_ee5cbe4f476b3dbda8d8f2dce0411466.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/835_ee5cbe4f476b3dbda8d8f2dce0411466.png)
In 40 s, the given athlete covers a distance of 200
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/835_ee5cbe4f476b3dbda8d8f2dce0411466.png)
In 1 s, the given athlete covers a distance =
![begin mathsize 12px style fraction numerator 200 straight pi over denominator 40 end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/4d59b3b9a01ad885b16dca44ec074897.png)
![begin mathsize 12px style fraction numerator 200 cross times 22 over denominator 40 cross times 7 end fraction cross times 140 end style](https://images.topperlearning.com/topper/tinymce/cache/2cf5d96d33a662409da78caa6850fccc.png)
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 120s his displacement is zero.
Then, the net displacement of the athlete is in 20 s only. In this interval of time, he moves at the opposite end of the 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
Distance covered by the athlete in 2 min 20 s is 2200 m and his displacement is 200 m.
Solution 2
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/Page_112_Q_2.jpg)
![begin mathsize 12px style Average space speed equals fraction numerator Total space distance space covered over denominator Total space time space taken end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/c2103eefcc1cb81e74fdf743fd9d9500.png)
![begin mathsize 12px style 300 over 150 end style](https://images.topperlearning.com/topper/tinymce/cache/4856e0979582a1fffcd74ff4e5f053cf.png)
![begin mathsize 12px style Average space velocity equals fraction numerator Displacement over denominator Time space interval end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/f0b35a3640dc3e0f8bd34227c7b22105.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/836_b0a24ad82c5fcd1d719d71dbc58efefd.png)
![begin mathsize 12px style 400 over 210 end style](https://images.topperlearning.com/topper/tinymce/cache/43973a7878b5efcf49aeadbfd53161ad.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/836_ffc9eeaee2fa9c470fffd309a6fa1f1c.png)
![begin mathsize 12px style 200 over 210 end style](https://images.topperlearning.com/topper/tinymce/cache/c827bc9007dcd959e7aae83b5b882b8c.png)
Solution 3
![begin mathsize 12px style Average space speed equals fraction numerator Total space distance over denominator Total space time space taken end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/e0d93ef395e229482e1c5d84239ff8bd.png)
![begin mathsize 12px style therefore 20 equals straight d over straight t subscript 1 end style](https://images.topperlearning.com/topper/tinymce/cache/1521c1d161015e996aba8961d0aa074b.png)
![begin mathsize 12px style straight t subscript 1 equals straight d over 20 end style](https://images.topperlearning.com/topper/tinymce/cache/84fb078c36f1e7d2984b94d291d64895.png)
![begin mathsize 12px style therefore 30 equals straight d over straight t subscript 2 end style](https://images.topperlearning.com/topper/tinymce/cache/567a9d53c2bdf89b4ce1997cd5cf693d.png)
![begin mathsize 12px style straight t subscript 2 equals straight d over 30 end style](https://images.topperlearning.com/topper/tinymce/cache/ca8282bbb0de2639594e4e7b7ed1ac05.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/837_1ae2aefd2fc6bcdb8d0ce81aa0ab626c.png)
![begin mathsize 12px style Average space speed equals fraction numerator 2 space straight d over denominator straight t subscript 1 plus straight t subscript 2 end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/cc2ffed418386ee0aa6fd9835ef79fac.png)
From equations (i) and (ii),
![begin mathsize 12px style Average space speed equals fraction numerator 2 space straight d over denominator begin display style straight d over 20 end style plus begin display style straight d over 30 end style end fraction equals fraction numerator 2 over denominator begin display style fraction numerator 3 plus 2 over denominator 60 end fraction end style end fraction equals 120 over 5 equals 24 space km divided by straight h end style](https://images.topperlearning.com/topper/tinymce/cache/e5aeddb878f5c7199b41e3a1302d0593.png)
Solution 4
Acceleration of the motorboat, a = 3 m/s2
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/838_df9bd03ce41d8cf2c5fef66eaf7e8d5a.png)
![begin mathsize 12px style straight s equals 0 plus 1 half cross times 3 cross times left parenthesis 8 right parenthesis squared
straight s equals 96 space straight m end style](https://images.topperlearning.com/topper/tinymce/cache/0e45661ac4da82b7a09259e01b8031c7.png)
Solution 5
For first car:
Initial speed of the car, u1 = 52 km/h = = 14.4 m/s
Time taken to stop the car, t1 = 5 s
Final speed of the car becomes zero after 5s of application of brakes.
For second car:
Initial speed of the car, u2 = 3 km/h = = 0.83 m/s
Time taken to stop the car, t2 = 10 s
Final speed of the car becomes zero after 10 s of application of brakes.
Plot of the speed versus time graph for the two cars is shown in the following figure:
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/Speed-time_graph_of_car.jpg)
Distance covered by first car = Area under the graph line AB
= Area of triangle OAB
= = 36 m
Distance covered by second car = Area under the graph line CD
= Area of triangle OCD
= = 4.15 m
Area of triangle OAB > Area of triangle OCD
Thus, the distance covered by first car is greater than the distance covered by second car.
Hence, the car travelling with a speed of 52 km/h travelled farther after brakes were applied.
Solution 6
(b) The distance-time graphs of the three objects A, B and C never meet at a single point. Thus, they are never at the same point on the road.
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/840_4d0842a9b90ce186199d40028f940ff3.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/840_e7cbe2cbe16175b490e49bd0ae9c9e35.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/840_4860f2a8e1562d341a7888083d90f4a4.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/840_7bcd68808353e4afed93cf0da18d2a41.png)
Hence, B has travelled a distance of 5.143 km when it passes C.
Motion Exercise 86
Solution 7
Acceleration, a = 10 m/s2
Initial velocity, u = 0 (since the ball was initially at rest)
Final velocity of the ball with which it strikes the ground, v
According to the third equation of motion:
v 2 = u 2 + 2 as
v 2 = 0 + 2 (10) (20)
v = 20 m/s
Concept Insight - Choose the equation of motion wisely out of the three, to minimize the number of steps in calculations.
The ball will strike the ground with a velocity 20 m/s.
According to the first equation of motion:
v = u + at
20 = 0 + 10 (t)
t = 2s
Hence, the ball will strike the ground after 2s with a velocity of 20 m/s.
Solution 8
(a) The distance travelled by the car in the first 4 seconds is given by the area between the curve and the time axis from t = 0 to t = 4 s. This area has been shaded in the graph below.
Number of squares in the shaded part of the graph = 62
Concept Insight: While counting the number of squares in the shaded part of the graph, the squares which are half or more than half are counted as complete squares but the squares which are less than half are not counted.
On X-axis,
5 squares represent 2 s.
1 square represents s.
On Y-axis,
3 squares represent 2 m/s.
1 square represents m/s.
So, area of 1 square on the graph =
Area of the shaded region of the graph =
Therefore, the car travels a distance of 16.53 m in the first 4 seconds.
(b) The uniform motion of the car is represented by the part AB of the graph, which represents constant speed.
Solution 9
When a car is moving in a circular track, its acceleration is perpendicular to its direction of motion at each instant.
Solution 10
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/844_ee5cbe4f476b3dbda8d8f2dce0411466.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/844_ee5cbe4f476b3dbda8d8f2dce0411466.png)
Speed of an object moving in a circular orbit,
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/844_01d77f5785d19d6fdb89782f2a8074b5.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/844_a00aadd09f69955173fc7b43c051722c.png)
Hence, the speed of the artificial satellite is 3.07 km/s