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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)

$50$

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

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

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

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

$=50$

Option 1)

$40$

Option 2)

$25$

Option 3)

$100$

Option 4)

$50$

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