"TH CLASS : 9UNIT :- MOTION SUBJECT:- PHYSICSQuestion 1. An object has moved through a distance. Can it have zero displacement?If yes, support your answer with an example.Answer : Yes, anobject even after it has moved through a distance, it can have zerodisplacement. As we know distance is just length of the path an object has coveredirrespective of its direction or position with reference to certain point, where as the shortestdistance measured from the initial to the final position of an object is known as thedisplacement.For example, an object starts from point A and after covering a distance of say 50 meters,reaches at point B. Here after, it again moves back to point A.Here the distance covered by object is = AB + BA = 50 m + 50 m = 100 mwhere as displacement of object is= AB - BA = 50 m - 50 m = 0 mAs initial position of object is same as that of its final position hence its displacement, whichis distance measured from the initial to the final position,is zero.A >-----------50 m-------------> B<-----------50 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 20seconds from his initial position?Answer : Suppose, a farmer moves along the boundary of a square field of side 10 m in 40s as shown in the figure given below.= Perimeter of SquareDistance cover by the farmer as he moves from A to B tofieldC to D to A,along the boundary wall of square field= 4 x side of square field= 4 × 10 m= 40 m ? speed of farmer = 40 m/40 s= 1 m/s = Speed × TimeDistance covered by farmer in 2 minutes20 seconds= 1 m/s × [(2×60) s + 20 s]= 140 mNumber of round in covering 40 mof distance = 1 roundalong the boundary wall? Number of round in covering 140 m of distance = 1×140 /40 rounds along the boundary wall= 3.5 round1 = 3 / rounds2 Which means the farmer will be at point C just diagonally opposite of point A1 ? Relative Displacement of farmer from point A at the end of 3 / round will be = length of 2 ACwhich can be determined by the mathematical theorem as given below :2 2 AC = vAB + vBC2 2 = v10 + v102 =10 v2= 10 × 1.414 m= 14.14mQuestion 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 : Both of the statements are not true as(a) Displacement can be zero (b) Its magnitude is either less or equal to the distance travelled by the objectQuestion 1. Distinguish between speed and velocity.Answer : The speed of an object is the distance covered per unit time,and velocity is thedisplacement per unit time. The speed is a scalar quantity as it has just magnitude where asvelocity is a vector quantity as it has both direction as well as magnitude.The speed can bechanged by the distance travelled by a body in a particular time where as the velocity can bechanged by changing the object's speed, direction of motion or both.Question 2. Under what condition(s) is the magnitude of average velocity of an objectequal to its average speed?Answer : Themagnitude of average velocity of an object is equal to its average speed,only when it is moving in a straight line.Question 3. What does the odometer of an automobile measure?Answer : Odometer of an automobile measures the distance covered by an automobile. AllAutomobiles are fitted with Odometer. Earlier Odometer used to be mechanical device, nowa days we have electronic odometer.Question 4. What does the path of an object look like when it is in uniform motion?Answer : The path of an object looks like a straight line when it is in uniform motion.jQuestion 5. During an experiment, a signal from a spaceship reached the ground station infive minutes.Question 5. What was the distance of the spaceship from the ground station? The signal8 -1 travels at the speed of light, that is, 3 × 10 m sAnswer :Speed of Signal (v) = Speed of light8 -1 =3 × 10 msTime taken by Signal to reach= 5 minutesthe ground station (t)= 5 × 60 seconds= 300 secondsDistance between the spaceship and the ground station (S) = vt8 -1 = 3 × 10 m s × 300 m10 =9×10 m Question 1. When will you say a body is in (i) uniform acceleration? (ii) nonuniformacceleration?Answer : (i) A body is said to be in uniform acceleration if it travels in a straight line andits velocity increases or decreases by equal amounts in equal intervals of time(ii)A body is said to be in nonuniform acceleration if the rate of change of its velocity is notconstant .-1 -1 Question 2. A bus decreases its speed from 80 km h to 60 km h in 5 s.Find the acceleration of the bus.Answer :-1 = 80 km hInitial Speed of the Bus (u)-1 = (80 × 1000)/ (60 × 60)ms -1= 800/36 ms -1 = 60 km hFinalSpeed of the Bus (v)-1 = (60 × 1000)/ (60 × 60)ms -1 = 600/36 ms = 5 sTime in transition (t)The acceleration of the Bus (a) = (v-u) / t -2 = [(800/36) - (600/36)] / 5 ms -2= (-200/36) / 5 ms -2 = 5.55 / 5 ms -2 = 1.11 ms Question 3. A train starting from a railway station and moving with uniform-1 acceleration attains a speed 40 km h in 10 minutes. Find its acceleration.Answer :-1= 0 ms Initial Speed of the Train (u)-1 = 40 km hFinalSpeed of the Train (v)-1 = (40 × 1000)/ (60 × 60)ms -1 = 400/36 ms = 10 minutesTime in transition (t)= 10 × 60 s= 600 s= 600 s= 600 sThe acceleration of the Train (a) = (v-u) / t-2 = [(400/36) - 0] / 600 ms -2= (11.11) / 600 ms -2 = 0.0185 ms Question 1. What is the nature of the distance-time graphs for uniform and non- uniform motion of an object?Answer :(i) For uniform speed, a graph of distance travelled against time is a straight line and notinclined along the time axis, as shown in the figure below(iI) For uniform speed, a graph of distance travelled against time is a curve andas shownin the figure belowQuestion 2. What can you say about the motion of an object whose distance-timegraph is a straight line parallel to the time axis? Answer :Motion of an object whose distance-time graph is a straight line parallel to thetime axis is not moving at all and is in state of rest as shown in the figure below : Question 3. What can you say about the motion of an object if its speed-time graph isa straight line parallel to the time axis?Answer :The motion of an object if its speed-time graph is a straight line parallel to the timeaxis indicates that the object is moving with uniform speed.Question 4. What is the quantity which is measured by the area occupied below thevelocity-time graph?Answer :The area occupied below the velocity-time graph measures the distance coveredby the object. -2 Question 1. A bus starting from rest moves with a uniform acceleration of 0.1 m s for2 minutes.Find (a) the speed acquired, (b) the distance travelled.Answer :Here as given, Initial speed (u) = 0-2 Acceleration (a) =0.1 m sTime in transition (t) =2 minutes=2 × 60 seconds= 120 sWe know that Final speed = u + at-2 -1? (a) the speed acquired = 0 + 0.1 m s × 120 m s-1= (1/10)120ms-1 = 12 ms2 = ut + (1/2)atWe know that distance(s)2 ? (b) the distance travelled. = 0 ×120 + (1/2) × 0.1×(120)= 0 + (120 × 120) /2 × 10= 14400/20 = 720 m=720 m1 Question 2. A train is travelling at a speed of 90 km h Brakes are applied so as to-2 produce a uniform acceleration of -0.5 m s Find how far the train will go before it isbrought to rest.Answer : -1 Given Initial speed of train (u) =90 km h-1 = (90 1000) / (60×60) m s-1 = 25 m s1 Final speed of train (v) = 0 ms-2 Braking acceleration (a) = -0.5 m s2 2 We know 2as = v - u2 2 Or distance (s) =(v -u )/2a2 2 ? Distance covered by the train before it came to rest =(0 -25 )/(2 ×-0.5 )m= - (25 × 25)×10/-1 m=625 mQuestion 3. A trolley, while going down an inclined plane, has an acceleration of 2 cm-2 s . What will be its velocity 3 s after the start?Answer : -1 Initial Velocity of trolley (u) =0 cms -2 Acceleration (a) = 2 cm sTime (t) = 3 sWe know that final velocity (v) = u + at-1 = 0 + 2 x 3 cms-1 ? The velocity of train after 3 seconds = 6 cmsQuestion 4. A racing car has a uniform acceleration of 4 m s-2. What distance will itcover in 10 s after start?Answer :-1 Initial Velocity of the car (u) =0 ms-2 Acceleration (a) = 4 m sTime (t) = 10 s2 We know Distance (s) = ut + (1/2)at2 ? Distance covered by car in 10 second = 0 × 10 + (1/2) × 4 × 10= 0 + (1/2) × 4× 10 × 10 m= (1/2)× 400 m= 200 m- Question 5. A stone is thrown in a vertically upward direction with a velocity of 5 m s 1 -2If the acceleration of the stone during its motion is 10 m s in the downwarddirection, what will be the height attained by the stone and how much time will it taketo reach there?Answer: -1 Given Initial velocity of stone (u) =5 m s-2 Downward of negative Acceleration (a) = 10 m s2 2 We know that 2 (a)(s) = v - u2 2 ?Height attained by the stone (s) = (0 -5 )/ 2× (-10)m= -25/-20 m= 1.25 mAlso we know that final velocity (v) = u + ator Time (t) = (v-u)/a? Time (t) taken by stone to attain the height(s) = (0-5)/- 10 s= 0.5 sQuestion 1. An athlete completes one round of a circular track of diameter 200 m in40 s. What will be the distance covered and the displacement at the end of 2 minutes20 s? Distance covered by the athlete Answer : Diameter of circular track (D) = 200 m Radius of circular track (r) = 200/2=100 mTime taken by the athlete for one round(t) = 40 sDistance covered by athlete in one round (s) = 2p r= 2 × ( 22 / 7 )×100Speed of the athlete (v) = Distance / Time= (2 × 2200)/(7 × 40)= 4400 / 7 × 40? Distance covered in 2 minutes 20= Speed (s) × Time(t)seconds (s) or 140 s= 4400 / (7 × 40) × (2×60 + 20)= 4400 /( 7 × 40) × 140= 4400 × 140 /7 × 40=2200 mNumber of round in 40 s =1 roundNumber of round in 140 s =140/401 =3 /2 After taking start from position X,the athlete will be at postion Y after1 3 / rounds as shown in figure2 ? Hence Displacement of the athlete =XYwith respect to initial position at X= Diameter of circular track= 200 mQuestion 2. Joseph jogs from one end A to the other end B of a straight 300 m roadin 2 minutes 30 seconds and then turns around and jogs 100 m back to point C inanother 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 mTotal time taken = 2 × 60 + 30 s=150 s ? Average Speed from AB = Total Distance / Total Time-1 =300/150 m s-1 =2 m s-1 ? Velocity from AB =Displacement AB / Time = 300/150 m s-1 =2 m sTotal Distance covered from AC =AB + BC=300 + 200 mTotal time taken from A to C = Time taken for AB + Time taken for BC= (2×60+30)+60 s= 210 s?Average Speed from AC = Total Distance /Total Time-1 = 400 /210 m s-1 = 1.904 m sDisplacement (S) from A to C = AB - BC= 300-100 m= 200 mTime (t) taken for displacement from AC = 210 s?Velocity from AC = Displacement (s) / Time(t)-1 = 200 / 210 m s-1 = 0.952 m sQuestion 3. Abdul, while driving to school, computes the average speed for his trip to-1 be 20 km h On his return trip along the same route, there is less traffic and the-1 average speed is 30 km h What is the average speed for Abdul.s trip?Answer : Let us assume:The distance Abdul commutes while driving from Home to = SSchoolLet us assume time taken by Abdul to commutes this distance = t1 Distance Abdul commutes while driving from School to Home = SLet us assume time taken by Abdul to commutes this distance = t2 -1 Average speed from home to school v = 20 km h1av"