A graph of x versus t is shown in figure. Choose correct alternatives given below

Options

A. At `E`, the acceleration `a lt 0`
B. At B`B`, the acceleration `a gt 0`
C. Average velocity for the motion between `A` and `D` is negative
D. The speed at `'D'` is more than that at `'E'`

Answer

A, C, D

Explanation

At `E`, the accleration `a lt 0`
At `C`, the slope of `x - t ` curve almost zero i.e., velocity at `C` is also zero.
At point `B` velocity `= 0`, before `B` veloctiy is positive so speed of particle is decreasing i.e., if `v gt 0` then
`a lt 0` (acceleration is negative hence at `B. a lt 0`).
`lt overset(vec)v gt` between `A ` and `D` is negtive because initial because is greater than final position.
From graph it is claer that
|slope at `D`| `gt` |Slope at `E`|