JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 11)
The position vector of a particle changes with
time according to the relation
$$\overrightarrow r (t) = 15{t^2}\widehat i + (4 - 20{t^2})\widehat j$$
What is the magnitude of the acceleration at t = 1 ?
What is the magnitude of the acceleration at t = 1 ?
50
25
40
100
Explanation
$$\overrightarrow r = \left( {15{t^2}} \right)\widehat i + \left( {4 - 20{t^2}} \right)\widehat j$$
$$\overrightarrow v = {{d\overrightarrow r } \over {dt}} = \left( {30t} \right)\widehat i - \left( {40t} \right)\widehat j$$
$$\overrightarrow a = {{d\overrightarrow v } \over {dt}} = \left( {30} \right)\widehat i - \left( {40} \right)\widehat j$$
$$\left| {\overrightarrow a } \right| = 50$$
$$\overrightarrow v = {{d\overrightarrow r } \over {dt}} = \left( {30t} \right)\widehat i - \left( {40t} \right)\widehat j$$
$$\overrightarrow a = {{d\overrightarrow v } \over {dt}} = \left( {30} \right)\widehat i - \left( {40} \right)\widehat j$$
$$\left| {\overrightarrow a } \right| = 50$$
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