The speed-time graph of a particle moving along a fixed direction is shown in the figure. Obtain the distance traversed by the particle between

(a) t = 0 s to 10 s,

(b) t = 2 s to 6 s.

Asked by Abhisek | 1 year ago |  159

##### Solution :-

(a) Distance traversed by the particle between t = 0 s and t = 10 s

= area of the triangle = ($$\dfrac{1}{2}$$) x base x height

= ($$\dfrac{1}{2}$$) x 10 x x12 = 60 m

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

(b) The 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 by the particle in time 5 to 6 s.

For the motion from 0 sec to 5 sec

Now, u = 0 , t = 5 , v = 12 m/s

From the equation v = u + at we get

a = $$\dfrac{ (v – u)}{t}$$$$\dfrac{12}{5}$$ = 2.4 m/s2

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

= ($$\dfrac{1}{2}$$) a (5)2 – ($$\dfrac{1}{2}$$) a (2)2 = ($$\dfrac{1}{2}$$) x 2.4 x (25 – 4) = 1.2 x 21 = 25.2 m

For the motion from 5 sec to 10 sec , u = 12 m/s and a = -2.4 m/s2

and t = 5 sec to t = 6 sec means n = 1 for this motion

Distance covered in the 6 the sec is S2 = u + ($$\dfrac{1}{2}$$) a (2n – 1)

= 12 – ($$\dfrac{2.4}{2}$$) (2 x 1 – 1) = 10.8 m

Therefore, the total distance covered from t = 2 s to 6 s = S1 + S2

= 25.2 + 10.8 = 36 m

Answered by Pragya Singh | 1 year ago

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