NCERT Solutions Class 9th Science Chapter – 7 Motion
| Textbook | NCERT |
| Class | 9th |
| Subject | Science |
| Chapter | 7th |
| Chapter Name | Motion |
| Category | Class 9th Science |
| Medium | English |
| Source | Last Doubt |
| NCERT Solutions Class 9th Science Chapter -7 Motion Question & Answer were prepared based on the latest exam pattern. We have Provided Is Matter Around to help students understand the concept very well. What is a motion in physics? What is motion answers? What is motion and its types? What is motion class 7? What is the SI unit of motion? Who found motion? Who defined motion first? What are these motions called? What are the properties of motion? What are the 4 main types of motion? |
NCERT Solutions Class 9th Science Chapter – 7 Motion
Chapter – 7
Motion
Question & Answer
Page: 74
| Question 1: An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example. Answer- Yes. An object that has moved through a distance can have zero displacement. Displacement is the shortest measurable 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 in a square park of length 20 m (as shown in the following figure). He starts walking from point A and after moving along all the corners of the park (point B, C, D), 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. |
| 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? Answer- The farmer takes 40 s to cover 4 × 10 = 40 m. In 2 min and 20 s (140 s), he will cover a distance Therefore, the farmer completes rounds (3 complete rounds and a half round) of the field in 2 min and 20 s. 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 = √(10)2 + (10)2 =√100 + 100 =√200 =10√2 =10 × 1.414 = 14.14 m 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. |
| 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. Answer- (a) Not true -Displacement can become zero when the initial and final position of the object is the same. (b) Not true – Displacement is the shortest measurable distance between the initial and final positions of an object. It cannot be greater than the magnitude of the distance travelled by an object. However, sometimes, it may be equal to the distance travelled by the object. |
Page: 76
| Question 1: Distinguish between speed and velocity. Answer-
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| Question 2: Under what condition(s) is the magnitude of average velocity of an object equal to its average speed? Answer- Average speed =Total distance covered\Total time taken Average velocity = Total Displacement covered\ Total time taken 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. |
| Question 3: What does the odometer of an automobile measure? Answer- The odometer of an automobile measures the distance covered by an automobile. |
| Question 4: What does the path of an object look like when it is in uniform motion? Answer- An object having uniform motion has a straight line path. |
| 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 × 108 ms−1. Answer- Time taken by the signal to reach the ground station from the spaceship = 5 min = 5 × 60 = 300 s Speed of the signal = 3 × 108 m/s Speed = Distance travelled\ Time taken ∴Distance travelled = Speed × Time taken = 3 × 108 × 300 = 9 × 1010 m Hence, the distance of the spaceship from the ground station is 9 × 1010 m. |
Page :77
| Question 1: When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration? Answer- (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 a body increases or decreases by equal amounts in an equal interval of time. (ii) 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 kmh−1 in 5 s. Find the acceleration of the bus. Answer- Initial speed of the bus, u = 80 km/h =80× 5\18 =22.22 m\s Final speed of the bus, v = 60 km/h =60× 5\18 =16.66 m\s Time take to decrease the speed, t = 5 s Acceleration, a= v-u\t =16.66 – 22.22 \5 = -1.112 m\s2 Here, the negative sign of acceleration indicates that the velocity of the car is decrea |
| Question 3: A train starting from a railway station and moving with uniform acceleration attains a speed 40 kmh−1 in 10 minutes. Find its acceleration. Answer- Initial velocity of the train, u = 0 (since the train is initially at rest) Final velocity of the train, v = 40 km/h Time taken, t = 10 min = 10 × 60 = 600 s a = v − u/t = 11.1−0/600 = 0.0185 ms−2 Hence, the acceleration of the train is 0.0185 m/s2. |
Page: 81
| Question 1: What is the nature of the distance−time graphs for uniform and non-uniform motion of an object? Answer- The distance−time graph for uniform motion of an object is a straight line (as shown in the following figure).  The distance−time graph for non-uniform motion of an object is a curved line (as shown in the given figure).  |
| 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? Answer- When an object is at rest, its distance−time graph is a straight line parallel to the time axis.  A straight line parallel to the x-axis in a distance−-time graph indicates that with a change in time, there is no change in the position of the object. Thus, the object 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? Answer-  Object is moving uniformly. A straight line parallel to the time axis in a speed−time graph indicates that with a change in time, there is no change in the speed of the object. This indicates the uniform motion of the object. |
Question 4: What is the quantity which is measured by the area occupied below the velocity−time graph? Answer- Distance The graph shows the velocity−time graph of a uniformly moving body. Let the velocity of the body at time (t) be v. Area of the shaded region = length × breath Where, Length = t Breath = v Area = vt = velocity × time …(i) We know, Velocity = DisplacementTIme Velocity = DisplacementTIme ∴ Displacement = Velocity × Time…(ii) From equations (i) and (ii), Area = Displacement Hence, the area occupied below the velocity−time graph measures the displacement covered by the body. |
Page: 82
| Question 1: A bus starting from rest moves with a uniform acceleration of 0.1 ms−2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled. Answer- (a) 12 m/s (b) 720 m (a) Initial speed of the bus, u = 0 (since the bus is initially at rest) Acceleration, a = 0.1 m/s2 Time taken, t = 2 minutes = 120 s Let v be the final speed acquired by the bus. ∴ a =v-u \t 0.1 = v-0\120 ∴v = 12 m/s (b) According to the third equation of motion: v2 − u2 = 2as Where, s is the distance covered by the bus (12)2 − (0)2 = 2(0.1) s 144 – 0 = 0.2s 144/0.2 = s s = 720 m Speed acquired by the bus is 12 m/s. Distance travelled by the bus is 720 m. |
| Question 2: A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of −0.5 ms−2. Find how far the train will go before it is brought to rest. Answer- Initial speed of the train, u = 90 km/h = 25 m/s Final speed of the train, v = 0 (finally the train comes to rest) Acceleration = −0.5 m s−2 According to third equation of motion: v2 = u2 + 2 ass=(25)2 \ 2(0.5)= 625 m (0)2 = (25)2 + 2 (− 0.5) sWhere, s is the distance covered by the train The train will cover a distance of 625 m before it comes to rest. |
| 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? Answer- Initial velocity of the trolley, u = 0 (since the trolley was initially at rest) Acceleration, a = 2 cm s−2 = 0.02 m/s2 Time, t = 3 s According to the first equation of motion: v = u + at Where, v is the velocity of the trolley after 3 s from start v = 0 + 0.02 × 3 = 0.06 m/s Hence, the velocity of the trolley after 3 s from start is 0.06 m/s. |
| Question 4: A racing car has a uniform acceleration of 4 ms−2. What distance will it cover in 10 s after start? Answer- Initial velocity of the racing car, u = 0 (since the racing car is initially at rest) Acceleration, a = 4 m/s2 Time taken, t = 10 ss= ut+1\2 ar2 According to the second equation of motion: Where, s is the distance covered by s=0+1\2×4×(10)2 =400\2 =200m the racing car Hence, the distance covered by the racing car after 10 s from start is 200 m. |
| Question 5: A stone is thrown in a vertically upward direction with a velocity of 5 ms−1. If the acceleration of the stone during its motion is 10 ms−2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there? Answer- Initially, velocity of the stone,u = 5 m/s 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. 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∴ t= -5\-10 =0.5s 0 = 5 + (−10) t According to the third equation of motion: v2 = u2 + 2 as s=52 \20 =1.25m (0)2 = (5)2 + 2(−10) s Hence, the stone attains a height of 1.25 m in 0.5 s. |
Page: 85
| 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 20s? Answer- Diameter of circular track (D) = 200 m Radius of circular track (r) = 200 / 2=100 m Time taken by the athlete for one round (t) = 40 s Distance covered by athlete in one round (s) = 2π r = 2 × ( 22 / 7 ) × 100 Speed of the athlete (v) = Distance / Time = (2 × 2200) / (7 × 40) = 4400 / 7 × 40 Therefore, Distance covered in 140 s = Speed (s) × Time(t) = 4400 / (7 × 40) × (2 × 60 + 20) = 4400 / (7 × 40) × 140 = 4400 × 140 /7 × 40 = 2200 mNumber of round in 40 s =1 round Number of round in 140 s =140/40 =3 1/2 After taking start from position X,the athlete will be at postion Y after 3 1/2 rounds as shown in figure  Hence, Displacement of the athlete with respect to initial position at x = xy = Diameter of circular track = 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? Answer- Total Distance covered from AB = 300 m Total time taken = 2 × 60 + 30 s =150 s  Therefore, Average Speed from AB = Total Distance / Total Time =300 / 150 m s-1 = 2 m s-1 Therefore, Velocity from AB =Displacement AB / Time = 300 / 150 m s-1 = 2 m s-1 Total Distance covered from AC =AB + BC =300 + 200 m Total time taken from A to C = Time taken for AB + Time taken for BC = (2 × 60+30)+60 s = 210 s Therefore, Average Speed from AC = Total Distance /Total Time = 400 /210 m s-1 = 1.904 m s-1 Displacement (S) from A to C = AB – BC = 300-100 m = 200 m Time (t) taken for displacement from AC = 210 s Therefore, Velocity from AC = Displacement (s) / Time (t) = 200 / 210 m s-1 = 0.952 m s-1 |
| Question 3. Abdul, while driving to school, computes the average speed for his trip to be 20 kmh−1. On his return trip along the same route, there is less traffic and the average speed is 40 kmh−1. What is the average speed for Abdul’s trip? Answer- 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 = Total distance Covered / Total time taken From equations (i) and (ii), Average Speed = 2d/d/12 2d x 12/d = 24 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 ms−2 for 8.0 s. How far does the boat travel during this time? Answer- Given Initial velocity of motorboat, u = 0 Acceleration of motorboat, a = 3.0 ms-2 Time under consideration, t = 8.0 s We know that Distance, s = ut + (1/2) at2 Therefore, The distance travel by motorboat = 0 ×8 + (1/2)3.0 × 82 = (1/2) × 3 × 8 × 8 m = 96 m |
| Question 5. A driver of a car travelling at 52 kmh−1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 kmh−1 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 travelled farther after the brakes were applied? Answer- As given in the figure below PR and SQ are the Speed-time graph for given two cars with initial speeds 52 kmh−1 and 3 kmh−1 respectively.  Distance Travelled by first car before coming to rest = Area of △ OPR = (1/2) × OR × OP = (1/2) × 5s × 52 kmh−1 = (1/2) × 5 × (52 × 1000) / 3600) m = (1/2) × 5 × (130 / 9) m = 325 / 9 m = 36.11 m Distance Travelled by second car before coming to rest = Area of △ OSQ = (1/2) × OQ × OS = (1/2) × 10 s × 3 kmh−1 = (1/2) × 10 × (3 × 1000) / 3600) m = (1/2) × 10 x (5/6) m = 5 × (5/6) m = 25/6 m = 4.16 m |
Question 6. Fig 8.11 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? Answer- (a) Object B (b) No (c) 5.714 km (d) 5.143 km Speed =Distance\Time (a) Slope of graph = y- axis \x- axis = Distance\ Time Therefore, Speed = slope of the graph Since slope of object B is greater than objects A and C, it is travelling the fastest. (b) All three objects A, B and C never meet at a single point. Thus, they were never at the same point on road. (c)  7 square box = 4 km ∴ 1 square box = 4/7 km C is 4 blocks away from origin therefore initial distance of C from origin = 16/7 km Distance of C from origin when B passes A = 8 km Thus, Distance travelled by C when B passes A = 8 – 16/7 = (56 – 16)/7 = 40/7 = 5.714 km (d)  Distance travelled by B by the time it passes C = 9 square boxes 9×4/7 = 36/7 = 5.143 km |
| 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? Answer- Let us assume, the final velocity with which ball will strike the ground be ‘v’ and time it takes to strike the ground be ‘t’ Initial Velocity of ball, u = 0 Distance or height of fall, s = 20 m Downward acceleration, a =10 m s-2 As we know, 2as = v2– u2 v2 = 2as+ u2 = 2 x 10 x 20 + 0 = 400 ∴ Final velocity of ball, v = 20 ms-1 t = (v – u)/a ∴Time taken by the ball to strike = (20 – 0)/10 = 20/10 = 2 seconds |
Question 8. The speed-time graph for a car is shown is Fig. 8.12. (a) Find out how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period. Answer-  The shaded area which is equal to 1/2 × 4 × 6 = 12 m represents the distance travelled by the car in the first 4 s. (b) Which part of the graph represents uniform motion of the car? Answer-  The part of the graph in red colour between time 6 s to 10 s represents uniform motion of the car. |
| 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. Answer- Possible When a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s2. (b) an object moving in a certain direction with an acceleration in the perpendicular direction. Answer- Possible When a car is moving in a circular track, its acceleration is perpendicular to its direction. |
| 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. Answer- Radius of the circular orbit, r = 42250 km Time taken to revolve around the earth, t = 24 h Speed of a circular moving object, v = (2π r)/t = [2× (22/7)×42250 × 1000] / (24 × 60 × 60) = (2×22×42250×1000) / (7 ×24 × 60 × 60) m s-1 = 3073.74 m s -1 |
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