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In the question number 35, the acceleration of the particle at `t = 1` isA. `2hatjm s^(-2)`B. `-2 hatj ms^(-2)`C. `4 hatj ms^(-2)`D. `-4hatj ms^(-2)` |
Answer» Correct Answer - C From the previous question, `vecv = 3hati + 4t hatj ms^(-1)` `therefore ` Acceleration, `veca = (dvecv)/(dt) = 4hatj ms^(-2)` Acceleration of the particle remains constant at all times. |
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