What can you say about the parallel to the time axis time graph is a straight line motion of an object if its speed?

• Answer:

  Speed-time graph is a straight line parallel to time axis, it means that the speed of the object is not changing with time i.e., the object is performing uniform motion.

  
  
  

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Page 2

• Answer:

  Area under the velocity-time graph gives the magnitude of displacement.
  

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Page 3

• Answer:

  Here, acceleration a = 0.1 ms-2 Time t = 2min = 2 x 60 = 120s Initial speed u = 0 (a) From Ist equation of motion, speed acquired, v = u + at = 0 + 0.1 x 120  = 12 ms-2 (b) From IInd equation of motion, Distance travelled,

  
  = 0.1 x 60 x 120 = 720 m
  

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Page 4

• Answer:

  Here, initial speed, u = 90 kmh-1 =  

  
   = 25 ms-1 Acceleration, a = - 0.5 ms-2 Train brought to rest, so final speed, v = 0 From third equation of motion, v2 = u2 + 2as 0 = (25)2 + 2 x 0.5 x s 0 = 625 - s s = 625 m
  

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Page 5

• Answer:

  Here, initial velocity, u = 0 Acceleration, a = 2 cm s-2 Time, t = 3 s From, Ist equation of motion, v = u + at = 0 + 2 x 3 = 6 ms-1
  

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Page 6

• Answer:

  
  
  
  
  
  
  
  The distance covered in 10 s by the car is 200 m.
  

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Page 7

• Answer:

  
  
  
  
  
  (i)
  
                        (ii)
  
  
                     
  
  
                            
  
  
  
              
  
  
   
  
                                 
  
  
  

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Page 8

• Answer:

  (c) Given, after half the circle, the particle will reach the diametrically opposite point i.e., from point A to point B. And we know displacement is shortest path between initial and final point.

  
  Displacement after half circle =AB=OA+OB           [
  
   Given, OA and OB = r]
  
  Hence, the displacement after half circle is 2r.
  

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Page 9

• Answer:

    (b) Given, initial velocity = u, height = h and a = g (acceleration due to gravity) At the highest point, final velocity becomes zero i.e., v = 0 From, third equation of motion,

  
                                                 
  
                                                 
  
                               
  
                           
  
  Here, we have used negative sign because the body is moving against the gravity.
  

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Page 10

• Answer:

    (d) Displacement of an object can be less than or equal to the distance covered by the object, because the magnitude of displacement is not equal to distance. However, it can be so if the motion is along a straight line without any change in direction. So, the ratio of displacement to distance is always equal to or less than 1.

  
  
  

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Page 11

• Answer:

    (b) From second equation of motion,

  
   If object starts from rest i.e., initial velocity (u) = 0 and aconite an acceleration (a) in time (t) Then,    
  
  
  
   if a = constant So, the object moves with constant or uniform acceleration.
  

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Page 12

• Answer:

    (a) From the given v-t graph, it is clear that the velocity of the object is not changing with time i.e., the object is in uniform motion.
  

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Page 13

• Answer:

    (c) In merry-go-round, the speed is constant but velocity is not constant, because its direction goes on changing i.e., there is acceleration in the motion. So, we can say that the boy is in accelerated motion.

  
   
  

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Page 14

• Answer:

    (b) Area under v-t graph represent displacement whose unit is metre or (m). Because, unit of velocity v = m/s and unit of time (T) = s.

  
             Unit of (v-t) graph
  
  . Hence, the unit of (v-t) graph is metre (m).
  

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Page 15

• Answer:

    (b) The slope of distance-time graph represents the speed. From the graph, it is clear that the slope of distance-time graph for car B is less than all other cars. So, the slope is minimum for car 6. Hence, car 6 is the slowest.
  

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Page 16

• Answer:

    (a) For uniform motion, the distance-time graph is a straight line (because in uniform motion object covers equal distance in equal interval of time).
  

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Page 17

• Answer:

    (c) Slope of velocity-time graph gives acceleration. Because slope of the curve

  
  where
  
   acceleration.
  

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Page 18

• Answer:

    (a) The distance moved and magnitude of displacement are equal only in the case of motion along a straight line. Because displacement is the shortest path between initial and find path. So, for car moving on straight road, distance moved and magnitude of displacement are equal.
  

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Page 19

• Answer:

    The displacement of a moving object in a given interval is zero i.e., the object comes back to its initial position in the given time (displacement is the shortest distance between the initial and final position of an object). The distance in this case will not be zero because distance is the total length of the path travelled by the body. If the object comes back to its initial position, then length of path travelled is not zero.
  

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Page 20

• Answer:

    We know that, the equations of uniformly accelerated motion are (i)

  
                         (ii)
  
                                 (iii)
  
  where, u = Initial velocity v = Final velocity a = Acceleration t = Time and    s = Distance For an object moving with uniform velocity (velocity which is not changing with time), then acceleration a = 0. So, equations of motion will become (putting a = 0 in above equations) (i)
  
                                    (ii)
  
                               (iii)
  
  
  

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Page 21

• Answer:

    From the graph, (i) Initial velocity,      u = 0                      [Since, displacement and time is zero] (ii) Velocity after 50s,

  
     [
  
  Given, displacement = 100m ]
  
                                           
  
  (iii) Velocity after
  
     [Here, displacement = zero and time = 100s]
  
  
  Therefore, Velocity-time graph plotted from the above data is shown below
  
   
  

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Page 22

Answer:

  Given, the car starts from rest, so its initial velocity u = 0 Acceleration, (a)

and time (t) = 8 s From first equation of motion, v = u + at On putting a

and t = 8 s in above equation, we get

So, final velocity v is

Again, from second equation of motion,

 On putting t = 8 s and a = 5 ms"2 in above equation, we get

So, the distance covered in 8 s is 160 m. Given,                                   total time t = 12 s. After 8 s, the car continues with constant velocity i.e., the car will move with a velocity of

. So,     remaining time t'= 12 s - 8 s = 4 s The distance covered in the last 4s (s') = Velocity x Time [

 Distance = Velocity x Time] = 40 x 4 = 160 m [We have used the direct formula because after 8 s, car is moving with constant velocity i.e., zero acceleration].

Total distance travelled in 12 s from the start D = s + s' = 160 + 160 = 320m

Page 23

• Answer:

    Let the distance between A and 6 be -x- km. Time taken in driving from A to B

  
     
  
  Similarly, time taken in returning from 6 to A.
  
  Average speed
  
  
  
  Hence, average speed of a motorcyclist is
  
  
  

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Page 24

• Answer:

    (i) From the graph, it is clear that velocity is not changing with time i.e., acceleration is zero.

  
  (ii) Again from the graph, we can see that there is no change in the velocity with time, so velocity after 15 s will remain same as
  
  (iii) Distance covered in 15s= Velocity x Time =20 x 15= 300m  
  
  
  

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Page 25

• Answer:

    When a stone is thrown vertically upwards, it has some initial velocity (let u). As the stone goes its velocity goes on decreasing (

  
  it is moving against the gravity) and at the highest point i.e., maximum height) its velocity become zero. Let the stone takes time 'f second to reach at the highest point. After that stone begins to fall (with zero initial velocity) and its velocity goes on increasing (since it is moving with the gravity) and it reaches its initial point of projection with the velocity v in the same time (with which it was thrown), So,

  Velocity	u	0	-u
  Time	0	t	2t

  Here, we have taken -u because in the upward motion velocity of stone is in upward direction and in the downward motion, the velocity is in downward direction. The velocity-time graph for the whole journey is shown below
  
  
  

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Page 26

• Answer:

    For first object given, u = 0 (because object dropped from rest) and time (t) = 2s.      From second equation of motion, the distance covered by first object in 2s is   

  
   
  
    
  
  
  
  Height of first object from the ground after 2 s
  
   = 150m -20m = 130m For second object given, u=0 and time (t) = 2 s From second equation of motion, the distance covered by second object in 2 s is
  
  
  
  Height of second object from the ground after 2s then
  
   = 100m - 20m = 80m Now, difference in the height after 2
  
  = 130 - 80= 50 m The difference in heights of the objects will remain same with time- as both the objects have been dropped from rest and are falling with same acceleration i.e., (acceleration due to gravity).
  

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