What is the nature of the distance-time graphs for uniform and non-uniform motion of an object ?( 2marks?

Text Solution

Solution : When the motion is uniform, the distance-time graph is a straight line with some slope. <br> When the motion is non-uniform, the distance time graph is not a straight line. It can any curve.

• Answer:

  (i) For uniform motion of an object, its distance-time graph is a straight line with constant slope. (ii) For non-uniform motion of an object, its distance-time graph is a curved line with increasing or decreasing slope.
  

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Page 2

• Answer:

  Distance-time graph is parallel to time axis, it means that the distance of the object is not changing with time i.e., the object is at rest.

  
  
  

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Page 3

• 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 4

• Answer:

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

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Page 5

• 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 6

• 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 7

• 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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• Answer:

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

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• Answer:

  
  
  
  
  
  (i)
  
                        (ii)
  
  
                     
  
  
                            
  
  
  
              
  
  
   
  
                                 
  
  
  

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Page 10

• 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 11

• 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 12

• 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 13

• 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 14

• 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 15

• 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 16

• 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 17

• 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 18

• 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 19

• Answer:

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

  
  where
  
   acceleration.
  

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Page 20

• 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 21

• 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 22

• 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 23

• 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 24

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 25

• 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 26

• 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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