NCERT Solutions Class 9 Science Chapter 8 Motion PDF

NCERT solutions for class 9 science chapter 8 Motion Topic 8.1 Describing Motions Q 1. An object has moved through a distance. Can it have zero displacements? If yes, support your answer with an example. e Answer:...

NCERT solutions for class 9 science chapter 8 Motion Topic 8.1 Describing Motions Q 1. An object has moved through a distance. Can it have zero displacements? If yes, support your answer with an example. e Answer: ut Yes, an object has moved through a distance can have zero displacements. tit If an object moves and returns to the original position, the displacement will be zero. Consider the movement in a circular path. A man walks from point A in a circular path in a park and comes back to point A. s In The distance traveled is equal to the circumference of the circular path, but displacement is zero. sh Q 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? ka Answer: Aa Side of the square field = The perimeter of the square = According to question, He completes 1 round in. Speed of the farmer = Distance covered in = Now, Number of round trips completed traveling = We know, in 3 round trips the displacement will be zero. In round, the farmer will reach diametrically opposite to his initial position. Displacement = e Q 3. Which of the following is true for displacement? ut (a) It cannot be zero. tit (b) Its magnitude is greater than the distance traveled by the object. Answer: s In (a) The first statement is false. Because displacement can be zero when the initial point coincides with the final point. sh (b) The second statement is false. The magnitude of displacement can never be greater than the distance travelled by the object. It can be either equal or less. Solutions of NCERT for class 9 science chapter 8 Motion ka Topic 8.2 Measuring the Rate of Motion Aa Q 1. Distinguish between speed and velocity. Answer: Speed Velocity  Speed is the distance Velocity is the speed of an object travelled by an object in unit moving in a definite direction. time Speed is a scalar quantity Velocity is a vector quantity e Speed does not depend on Velocity changes with change in ut the direction direction tit Speed can never be negative Velocity can be positive, negative or zero. s In Q 2. Under what condition(s) is the magnitude of the average velocity of an object equal sh to its average speed? Answer: ka When the total distance traveled by the object is equal to the displacement, the magnitude of the average velocity will be equal to the average speed. Average speed is the total distance upon the time taken, whereas average velocity is the total Aa displacement upon time taken. Q 3. What does the odometer of an automobile measure? Answer: Odometer is a device that measures the total distance traveled by automobile. Q 4. What does the path of an object look like when it is in uniform motion? Answer: An object is having a uniform motion if it covers equal distance in equal interval of time (which implies speed is constant!). So the path can be straight or curved. e For eg. Consider a circular path. For understanding purposes, divide the circumference ut of the circle in six equal parts each subtending at the centre. The object covers each equal part in equal amount of time. Hence, by definition, this object is in uniform tit motion. Q 5. During an experiment, a signal from a spaceship reached the ground station in five s minutes. What was the distance of the spaceship from the ground station? The signal In travels at the speed of light, that is ,. Answer: sh Given, the signal travels at the speed of light,. Time taken by the signal = ka Let the distance of the spaceship from the ground station be Aa We know, Therefore, the distance of spaceship from the ground station is NCERT textbook solutions for class 9 science chapter 8 Motion Topic 8.3 Rate of Change of Velocity e Q 1. When will you say a body is in ut (i) uniform acceleration? (ii) nonuniform acceleration? tit Answer: s (i) If the velocity of an object traveling in a straight line increases or decreases In by equal amounts inequal intervals of time, then the acceleration of the object is said to be uniform. For example, An apple having a free-fall motion. sh (ii) On the other hand, if the velocity of the object increases or decreases by unequal amounts inequal intervals of time, then the acceleration of the object is said to be non-uniform. For example, A car travelling along a straight road increasing its ka speed by unequal amounts in equal intervals of time. Q.2 A bus decreases its speed from 80 km h -1 to 60 km h -1 in 5 s. Find the acceleration Aa of the bus. Answer: (We know, ) Given, Initial speed of the bus, = The final speed of the bus, = Time is taken, We know, e ut The negative sign implies retardation. Therefore, the acceleration of the bus is tit Or, the retardation(de-acceleration) of the bus is s Q 3. A train starting from a railway station and moving with uniform acceleration attains In a speed 40 km h -1 in 10 minutes. Find its acceleration. Answer: sh (We know, ) Given, The train starts from rest. Hence, the initial speed of the train ka = Final speed of the train = Aa Time taken, We know, Therefore, the acceleration of the train is Solutions for NCERT class 9 science chapter 8 Motion Topic 8.4 Graphical Representation of Motion e Q 1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object? ut Answer: tit Distance-time graph is the plot of distance travelled by an object along x-axis against time along y-axis. s For the uniform motion of an object, the distance-time graph is a straight line with a In constant slope. (Note: However, the path may be straight or curved!) sh ka Aa For non-uniform motion of an object, the distance-time graph is a curved line with an increasing or decreasing slope.  e ut Q2. What can you say about the motion of an object whose distance-time graph is a tit straight line parallel to the time axis? Answer: s In If the distance-time graph of an object is a straight line parallel to the time axis, it means that the distance of the object is the same from its initial position at any point of time. This implies that the object is not moving and is at rest. sh Q3. What can you say about the motion of an object if its speedtime graph is a straight line parallel to the time axis? ka Answer: If the speed-time graph of an object is a straight line parallel to the time axis, it means Aa that the speed of the object is not changing with time. Hence the speed of the object is constant. This also implies that the acceleration of the object is zero. Q4. What is the quantity which is measured by the area occupied below the velocity- time graph?  Answer: The area occupied below the velocity-time graph denotes the total distance travelled by an object in the given time frame. We know, e ut s tit Topic 8.5 Equations of motion by graphical method In sh Q1.(a) A bus starting from rest moves with a uniform acceleration of 0.1 m s -2 for 2 minutes. Find ka the speed acquired Answer: Aa Given, The bus starts from rest. Hence, the initial speed of the bus = Acceleration of the bus, = Time is taken, (a) We know,  Therefore, the speed acquired by the bus is Q1.(b) A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find e The distance travelled. ut Answer: tit Given, The bus starts from rest. Hence, the initial speed of the bus, u = Acceleration of the bus, = s In Time taken, (b) We know, sh Therefore, the distance travelled by bus is ka Q2. 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 Aa rest. Answer: (We know, ) Given, Initial speed of the train, = Acceleration of the train, (Negative sign implies retardation) Since, the train has to be brought to rest, final speed of the train, = We know, e ut Therefore, the train travels a distance of before coming to rest. tit Q3. 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: s In Given, The trolley starts from rest. Hence, the initial speed of the trolley, = Acceleration of the trolley, = sh Time is taken, ka (a) We know, Aa Therefore, the velocity of the trolley after 3 sec is Q4. A racing car has a uniform acceleration of 4 m s -2. What distance will it cover in 10 s after start? Answer: Given, Initial speed of the racing car, u = e Acceleration of the car, = ut Time taken, tit We know, s In Therefore, the distance travelled by the racing car in is sh Q 5. A stone is thrown in a vertically upward direction with a velocity of. If the acceleration of the stone during its motion is in the downward direction, what will be the height attained by the stone and how much time will it take to reach there? ka Answer: Aa Taking upward direction as positive (+) direction: Given, (This is due to gravitational force!) The stone will move up until its velocity becomes zero. We know, e ut Therefore, the stone reaches to a height of tit Now, We know, s In Therefore, the time taken by the stone to reach the maximum height is. sh NCERT solutions for class 9 science chapter 8 Motion: Solved Exercise ka Questions- Q 1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What Aa will be the distance covered and the displacement at the end of 2 minutes 20 s? Answer: Given, Diameter of the circular track = The circumference of the circular track, The athlete completes one round of a circular track in 40 s. Speed of the athlete = In , e Distance travelled by the athlete = ut tit Also, number of rounds the athlete will complete in = Therefore, the final position of the athlete after will be diametrically opposite to his initial point. s In (3 complete rounds and one half round.) Hence, displacement of the athlete = magnitude of diameter of the circle =200m sh Q2. 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 ka another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and Aa (b) from A to C? Answer: Given, (a) Distance between A and B = Time taken to reach from A to B = Average speed from A to B = And, Average velocity from A to B = e (In this case, average speed is equal to the average velocity) ut (b) Distance travelled from A to reach C = tit And, Displacement from A to C = Also, time taken to reach C from A = s In Average speed from A to C = And, Average velocity from A to C = sh (In this case, average speed is not equal to the average velocity) Q 3. Abdul, while driving to school, computes the average speed for his trip to ka be. On his return trip along the same route, there is less traffic and the average speed is. What is the average speed for Abdul’s trip? Aa Answer: Given, Average speed while going to school, And Average speed while returning back from school, Let the distance between starting point and school be And time taken by Abdul during the two trips be We know, e And, -(i) ut Now, Total distance that Abdul covers = tit And total time Abdul takes to cover this distance = s In Therefore, the average speed for Abdul's trip is (Note: ) sh ka Q 4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of for 8.0 s. How far does the boat travel during this time? Aa Answer: Given, The motorboat starts from rest. Hence, initial speed of the motorboat, u = Acceleration of the motorboat, = Time taken, We know, e Therefore, the distance travelled by the motorboat is ut Q 5. A driver of a car travelling at 52 km h -1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h - tit 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 s farther after the brakes were applied? In Answer: The initial speed sh After 5 sec the car stops The graph is represented by the blue line ( x-axis is time and the y-axis is speed) ka For the car with 3Kmh -1. Initial speed. The graph which is Aa represented by the golden line  e ut s tit The area covered by the blue graph is greater than the golden graph so the car with 15 In m/s initial velocity travells large distance. Q 6. (a) Figure shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions: sh ka Aa Which of the three is travelling the fastest?  Answer: Given is a distance-time graph. The slope of this graph gives us speed. Hence, the graph with the highest slope will have the highest speed. Since B has the highest slope(inclination), it travels the fastest. e Q 6. (b) Figure shows the distance-time graph of three objects A,B and C. Study the ut graph and answer the following questions: s tit In sh Are all three ever at the same point on the road? ka Answer: Aa Given is a distance-time graph. Any point on the curve will give the distance of object from O. Since there is no intersection point of all the three graphs, they never meet at the same point on the road. (Although any two of them do meet at some point on the road!)  Q 6.(c) Figure shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions: e ut tit How far has C travelled when B passes A? Answer : s In Given is a distance-time graph. Any point on the curve will give the distance of object from O. To find how far C has travelled when B passes A, draw a perpendicular from sh the intersection point of A and B on time axis. The point where it intersects on the C graph, from C draw a perpandicular to y axis. Therefore, distance travelled by C will be (Final distance from O - Initial distance from O) ka Therefore, C has traveled 6.5 km when B passes A. Aa Q 6. (d) Figure shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:  e ut How far has B travelled by the time it passes C? tit Answer: s Given is a distance-time graph. The graph of B and C intersect at a point whose y- coordinate is 5. Hence, B has travelled by the time it passes C. In Q 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 sh strike the ground? Answer: ka Considering downward direction as positive direction. Given, Height from which ball is dropped, Aa Acceleration of the ball, = Initial velocity, (i) We know,  (In downward direction) Therefore, the ball will strike the ground with a velocity of (ii) Now, we know, e ut Therefore, the ball reaches the ground in. tit Note: was rejected because in this case, the negative sign implies the velocity in upward direction, which is opposite to the direction of the motion of the ball(before collision). s In Q 8.(a) The speed-time graph for a car is shown is Figure: sh ka 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. Aa Answer: Given is a speed-time graph. The area under the curve will give the distance travelled by the car.  In time , the distance travelled by the car will be equal to the area under the curve from Considering this part of the graph as a quarter of a circle whose radius = 4 unit. Therefore, required area = e Therefore, distance the car travelled in the first 4 seconds is ut Q 8. (b) The speed-time graph for a car is shown is Figure: s tit In Which part of the graph represents uniform motion of the car? sh Answer: In uniform motion, the speed of car will become constant which is represented by line ka parallel to the time axis. In the given figure, the straight line graph from represents the uniform motion of the car. Aa Q 9. (a) State which of the following situations are possible and give an example for each of these: an object with a constant acceleration but with zero velocity Answer:  (a) The given situation is possible. When an object is thrown upwards (under gravity only), it reaches to a maximum height where its velocity becomes zero. However, it still has an acceleration acting in the downward direction (acceleration due to gravity). e Note: This is possible for a given point of time, however, it is not possible for a period of ut time. Q 9.(b) State which of the following situations are possible and give an example for tit each of these: an object moving with an acceleration but with uniform speed. s In Answer: (b) The given situation is possible. sh An object moving in a circular path with uniform speed, i.e covering equal distance in equal amount of time is still under acceleration. Because, the velocity keeps on changing due to continuous change in the direction of motion. Therefore, circular motion ka is an example of an object moving with an acceleration but with uniform speed. Q 9. (c) State which of the following situations are possible and give an example for Aa each of these: an object moving in a certain direction with an acceleration in the perpendicular direction. Answer: (c) The given situation is possible. For an object moving in a circular path with constant speed, the direction of its velocity at any point will be tangential to that point. However, its acceleration will be directed radially inwards. (Constant speed but still having an acceleration - Due to continuous change in direction.) Q 10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its e speed if it takes 24 hours to revolve around the earth. ut Answer: tit Given, Radius of the circular orbit = Circumference of the orbit = s In The satellite takes 24 hours to revolve around the earth. We know, sh Therefore, the speed of the artificial satellite is ka Aa