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Help me please, - Kinematics - JEE Main

The position vector of a particle changes with time according to the relation \vec{r}(t)=15t^{2}\:\hat{i}+(4-20t^{2})\:\hat{j}. What is the magnitude of the accelaration at t=1 ?

  • Option 1)

    40

  • Option 2)

    25

  • Option 3)

    100

  • Option 4)

    50

Answers (1)
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S solutionqc

\vec{r}\left ( t \right )=15t^{2}\hat{i}+(4-20t^{2})j

\vec{v}(t)=\frac{d\vec{r}(t)}{dt}=30t\:\hat{i}-40\:t\:\hat{j }

\vec{a}(t)=\frac{d\vec{v}(t)}{dt}=30\:\hat{i}-40\:\:\hat{j }

\left | \vec{a} \right |=\sqrt{30^{2}+40^{2}}

=50

 

 

 


Option 1)

40

Option 2)

25

Option 3)

100

Option 4)

50

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