3.27 The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s  (b) t = 2 s to 6 s.

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    Vishal Baghel | Contributor-Level 10

    5 months ago

    3.27

    (a) Distance travelled by the particle between t = 0 s and t = 10 s is the area of the triangle = (1/2) x base x height = (1/2) x 10 x 12 = 60 m

    The average speed of the particle is 60/10 m/s = 6 m/s

     

    (b) Distance travelled by the particle between t = 2 s and t = 6 s

    Let S1 be the distance travelled by the particle in time 2 to 5 s and S2 be the distance travelled between 5 to 6s.

    For the motion  from 0 to 5 sec, u = 0, t = 5, v = 12 m/s

    From the equation v = u + at we get a = (v-u)/t = 12/5 = 2.4 m/s2

    Distance covered from 2 to 5 sec, S1 = distance covered in 5 s – distance covered in 2 s

    From the equation s = ut + at2 we

    ...more

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According to question, we can write

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

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Average speed = 4 v 2 3 v

= 4 v 3

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Final velocity = v i ˆ

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

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