The acceleration time graph of a particle is shown in the Fig. 2 (CF). 11. At time ` t= 10 s` is the particle is ` 8 ms^(-1)`. Its velocity ` t= 10 s` is.
.

Options

A. (a) (50)/3 ma^(-1)`
B. (b) (70)/3 ma^(-1)`
C. (c ) ` (74)/3 ms^(-1)`
D. (d) (144)/3 ms^(-1)`

Answer

C

Explanation

As, acceleration = (change in velocity)/(time taken)` ltbRgt :. Change in velocity=acceleration xx time taken
Hence, change in velocity in ` 10 s=area OAB`
`+area BCD`
` = 1/2 xx 6 xx 10 +1/2 xx (10 -6 ) xx (- (20)/3)`
`= 30 - (40)/3 = (50)/3 ms^(-1)`
:. Final velocity at ` 10 s=initail veloicty+change in velocity = 8 + (50)/3 = (74)/3 ma^(-1)`.