A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates with the constant rate and comes to rest. If the total time taken is 4 sec, the distance travelled is

Options

A. 32m
B. 16m
C. 4m
D. 14m

Answer

B

Explanation

`8= a t_(1) and 0 = 8 -a (4- t_(1)) or t_(1) = (8)/(a) :. 8=a (4-(8)/(a))`
`8=4a-8 or a= 4 and t_(1) = 8//4 = 2 sec`
Now `s_(1) = 0 xx 2 + (1)/(2) xx 4 (2)^(2) or s_(1) = 8m`
`s_(2) = 8 xx 2 - (1)/(2) xx 4 xx (2)^(2) or s_(2) = 8m`
`:. s_(1) + s_(2) = 16m`