The displacement of a particle is moving by `x = (t - 2)^2` where `x` is in metres and `t` in second. The distance covered by the particle in first `4` seconds is.

Options

A. 4 m
B. 8 m
C. 12 m
D. 16 m

Answer

B

Explanation

Given, `x=(t-2)^(2)`

Velocity, `upsilon=(dx)/(dt)=(d)/(dt)(t-2)^(2)=2(t-2)ms^(-1)`
Acceleration, `a=(d upsilon)/(dt)=(d)/(dt)[2(t-2)]=2(1-0)=2ms^(-1)`
When, `t=0, upsilon = -4ms^(-1), t=2s, upsilon=0 ms^(-1)`
`t=4 s, upsilon = 4ms^(-1)`
`upsilon-t` graph is shown in adjacent diagram.
Distance travelled = area of the graph
= |area OAC|+ area ABD
`|(4xx2)/(2)|+(1)/(2)xx2xx4=8m`