JEE Mains · Physics · STD 11 - Motion in plane
- JEE Mains 2019
- Medium
- Numerical
The position vector of a particle changes with time according to the relation \(\vec r\left( t \right) = 15{t^2}\hat i + \left( {4 - 20{t^2}} \right)\hat j\). What is the magnitude of the acceleration at \(t = 1\) ?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(50\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
\vec r = \left( {15{t^2}} \right)\hat i + \left( {4 - 20{t^2}} \right)\hat j\\
\vec v = \frac{{d\vec r}}{{dt}} = \left( {30t} \right)\hat i - \left( {40t} \right)\hat j\\
\vec a = \frac{{d\vec v}}{{dt}} = \left( {30} \right)\hat i - \left( {40} \right)\hat j\\
\left| {\vec a} \right| = 50
\end{array}\)
\vec r = \left( {15{t^2}} \right)\hat i + \left( {4 - 20{t^2}} \right)\hat j\\
\vec v = \frac{{d\vec r}}{{dt}} = \left( {30t} \right)\hat i - \left( {40t} \right)\hat j\\
\vec a = \frac{{d\vec v}}{{dt}} = \left( {30} \right)\hat i - \left( {40} \right)\hat j\\
\left| {\vec a} \right| = 50
\end{array}\)
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