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Answered on 15/09/2018 Learn Motion in a Straight Line +1 Physics

Mandeep Kaur

Distance travelled will be the area under the speed time curve. (a) Distance travelled is: 2 x 1/2 (5x12) = 60m (b) Slope of upward increasing curve: 12/5 and of decreasing curve it is: -12/5 This gives the rate of change of speed with time. Thus equation of line, speed at 2nd second... read more
Distance travelled will be the area under the speed time curve.
 
(a) Distance travelled is: 2 x 1/2 (5x12) = 60m
 
(b) Slope of upward increasing curve: 12/5 and of decreasing curve it is: -12/5
      This gives the rate of change of speed with time.
      Thus equation of line, speed at 2nd second will be:12/5*2 = 4.8 m/s
      Speed at 6th second: 12 - 12/5*1 = 48/5 = 9.6 m/s....  (because rate of change is negative
      and from 12 to 0)
      Thus total distance travelled = 1/2(12-4.8)(5-2) + 1/2(12-9.3)(6-5) = 12.15 m
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Answered on 14/09/2018 Learn Motion in a Straight Line +1 Physics

Abhishek Aman

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Both are in opposite direction,so consider one velocity in negative direction. Velocity of jet w.r.t. ground Vjg= 500 km/h …...(Upward) Velocity of products relative to jet VPJ = 1500 km/h ... read more

Both are in opposite direction,so consider one velocity in negative direction.

    Velocity of jet w.r.t. ground Vjg= 500 km/h                                                                …...(Upward)

        Velocity of products relative to jet VPJ = 1500 km/h                                                …(downward)

Hence, velocity of  products relative to ground = -1500+500

                                                                                    =  -1000 km/h

Here – ve sign means downward .

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Answered on 15/09/2018 Learn Motion in a Straight Line +1 Physics

Vijay Giri

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Distance Vs time graph is parabolic graph in case of n sec but it's straight line if we draw graph for distance Vs nth second time graph passing through origin as u=0 and acceleration constant
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Answered on 14/09/2018 Learn Motion in a Straight Line +1 Physics

Anurag Mohapatra

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Speed of the belt, vB = 4 km/h Speed of the boy, vb = 9 km/h Since the boy is running in the same direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as: vbB = vb + vB = 9 + 4 = 13 km/h Since the boy is running in the direction opposite to the direction... read more

Speed of the belt, vB = 4 km/h

Speed of the boy, vb = 9 km/h

Since the boy is running in the same direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:

vbB = vb + vB = 9 + 4 = 13 km/h

Since the boy is running in the direction opposite to the direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:

vbB = vb + (-vB) = 9 – 4 = 5 km/h

Distance between the child’s parents = 50 m

As both parents are standing on the moving belt, the speed of the child in either direction as observed by the parents will remain the same i.e., 9 km/h = 2.5 m/s.

Hence, the time taken by the child to move towards one of his parents is .

If the motion is viewed by any one of the parents, answers obtained in (a) and (b) get altered. This is because the child and his parents are standing on the same belt and hence, are equally affected by the motion of the belt. Therefore, for both parents (irrespective of the direction of motion) the speed of the child remains the same i.e., 9 km/h.

For this reason, it can be concluded that the time taken by the child to reach any one of his parents remains unaltered.

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Asked on 14/09/2018 Learn Motion in a Straight Line +1 Physics

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h-1 .Finding... read more

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h-1 .Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h-1 What is the (a)Magnitude of average velocity, and (b)Average speed of the man over the interval of time (i) 0 to 30 min. (ii) 0 to 50 min. (iii) 0 to 40 min? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]

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Asked on 14/09/2018 Learn Motion in a Straight Line +1 Physics

Figure shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph... read more

Figure shows the x-t plot of one-dimensional motion of a particle.
Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.
ncert-solutions-class-11th-physics-chapter-3-motion-straight-line-15

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