3.22 Figure 3.25 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D?

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    Vishal Baghel | Contributor-Level 10

    5 months ago

    3.22 The change in the speed with time is maximum in interval 2. Therefore, the average acceleration is the greatest in magnitude in interval 2

    The average speed is maximum in interval 3

    The sign of velocity is positive in intervals 1,2 and 3

    Acceleration depends on the slope. The acceleration is positive in interval 1 and 3, as the slope is positive. The acceleration is negative in interval 2, as the slope is negative

    Acceleration at A, B, C and D is zero since the slope is parallel to the time axis at these instants

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A
alok kumar singh

Kindly go through the solution '

 

A
alok kumar singh

Please find the solution below:

 

 

V
Vishal Baghel

[h] = ML2T-1

[E] = ML2T-2

[V] = ML2T-2C-1

[P] = MLT-1

 

A
alok kumar singh

According to question, we can write

 10 = 1 2 a t 2 . . . . . . . . . . . . . . . ( 1 ) a n d                        

1 0 + x = 1 2 a ( 2 t ) 2 . . . . . . . . . . . . . . . . ( 2 )  

X = 1 2 A [ ( 2 t ) 2 t 2 ] = 3 ( 1 2 a t 2 ) = 3 0 m  

 

A
alok kumar singh

Average speed = 4 v 2 3 v

= 4 v 3

(d) Initial velocity  = - v j ˆ

Final velocity = v i ˆ

Change in velocity  = v i ˆ - ( - v j ˆ )

= v ( i ˆ + j ˆ ) Momentum gain is along i ˆ + j ˆ

 Force experienced is along i ˆ + j ˆ

  Force experienced is in North-East direction.

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