Displacement time graph of a particle moving in a straight line is as shown in figure.
From the graph we can conclude that work done on the block is :
.

Options

A. positive from `0` to `t_(1)`
B. negative from `t_(1)` to `t_(2)`
C. zero from `t_(2)` to `t_(3)`
D. negative from `t_(3)` to `t_(4)`.

Answer

A, B, C

Explanation

Slope of the graph at a point = velocity,
`t_(1) = 0` to `t_(1)` velocity increases.
`t_(1)` to `t_(2)` velocity decreases, `t_(2)` to `t_(3)` velocity zero and
`t_(3)` to `t_(4)` velocity is constant.