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The graph between the displacement `x` and time `t` for a particle moving in a straight line is shown in the figure. During the interval `OA, AB, BC` and `CD` the acceleration of the particle is
`OA, AB, BC, CD`
image
A. `OA=+, AB=0, BC=+, CD=+`
B. `OA=-, AB=0, BC=+, CD=0`
C. `OA=+, AB=0, BC=-, CD=+`
D. `OA=-, AB=0, BC=0, CD=+`

1 Answer

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Best answer
Correct Answer - D
`a=(d^(2)x)/(dt^(2))=` change in velocity w.r.t the time
For `OA rarr` velocity decreases so a is negative
For `AB rarr` velocity constant so a is zero.
For `BC rarr` velocity constant so a is zero
For `CD rarr` velocity increases so a is positive.

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