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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` 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=+` |
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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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