A vehicle starts moving in a straight line with an acceleration, `a = 4 m//s^(2)`, with initial velocity equal to zero. After accelerating for time `t_(1)`, the vehicle moves uniformly and for time `t_(2)`, the vehicle finally decelerates for time, `t_(1)` eventually coming to a stop. The total time taken during the motion is 10s and the average velocity during the motion is 5.1 m/s. The time taken by the vehicle during acceleration is

Options

A. 2s
B. 2.5 s
C. 1.5 s
D. 1.8s

Answer

C

Explanation

For first part of journey, x = 0, a = 4 `ms^(-2)`
time =`t_(1)`. Velocity attained by particle at end of time `t_(1)` is `v_(1) = u + "at"_(1) = 4t_(1)`
`:.` Displacement in first `t_(1)` seconds is
`S_(1) = (1)/(2) "at"_(1)^(2) = (1)/(2) xx 4 xx t_(1)^(2) = 2t_(1)^(2)`
For second part of journey, initial velocity, `v_(1) =4t_(1)`, time =`t_(2)` and acceleration= 0
`:.` Displacement in next `t_(2)` second is `s_(2) = v_(1) t_(2)= 4t_(1) t_(2)`
For third part of journey,
Initial velocity, `v_(1) = 4t_(1)`
Time interval =`t_(1)`
So, acceleration `= (0-4t_(1))/(t_(1)) = -4 ms^(-2)`
`:.` Displacement in third part is
`s_(3) = (-16t_(1)^(2))/(-8) = 2t_(1)^(2) ( :. v_(2)^(2) - v_(1)^(2) = 2as_(3))`
Now, `v_("avg") = ("Total displacement")/("Total time")`
`5.1 = (s_(1) + s_(2) + s_(3))/(t_(1) + t_(2) + t_(1))` [ `:.` third time interval `=t_(1)`]
`5.1 = (2t_(1)^(2) + 4t_(1) t_(2) + 2t_(1)^(2))/(10)`
`rArr 51 = 4t_(1)^(2) + 4t_(1) (10-2t_(1))`
`rArr 51 = 4t_(1)^(2) + 40t_(1) - 8t_(1)^(2)`
`rArr 4t_(1)^(2) - 40t_(1) + 51 = 0`
`rArr 4t_(1)^(2) - 34t_(1) - 6t_(1) + 51 = 0`
`rArr (2t_(1) - 3) (2t_(1) - 17) = 0 rArr t_(1) = (3)/(2)`
or `t_(1) = (17)/(2) rArr t_(1) = 1.5s`
(As `t_(1) -8.5s` is not possible . It exceeds total time)