A particle moving along a straight line with a constant acceleration of `-4 m//s^2` passes through a point `A` on the line with a velocity of `+ 8 m//s` at some moment. Find the distance travelled by the particle in `5` seconds after that moment.

Options

A. 26 m
B. 18 m
C. 9 m
D. 32 m

Answer

A

Explanation

`u = +8 m//s, a= -4 m//s^2`
`v = 0`
`rArr 0= 8 - 4t`or `t = 2 sec`
Displacement in first `2 sec`,
`S_1 = 8 xx 2 + (1)/(2).(-4). 2^2 = 8 m`
Displacement in next `3 sec`,
`S_2 = 0 xx 3 + (1)/(2)(-4) 3^2 = -18 m`.
distance travelled `= |S_1| + |S_2| = 26 m`
ALITER: We can draw velocity time graph of the situation. total distance is equal to magnitude of area under velocity time graph
Hence
`d =(1)/(2) xx 2 xx 8 + (1)/(2) xx 3 xx 12 = 8 + 18 = 26 m`.