MCQ
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 $t = 1$ ?
- A$40$
- B$100$
- C$25$
- ✓$50$
✓
Answer
Correct option: D.
$50$
d
$\begin{array}{l}
\vec r = \left( {15{t^2}} \right)\hat i + \left( {4 - 20{t^2}} \right)\hat j\\
\vec v = \frac{{d\vec r}}{{dt}} = \left( {30t} \right)\hat i - \left( {40t} \right)\hat j\\
\vec a = \frac{{d\vec v}}{{dt}} = \left( {30} \right)\hat i - \left( {40} \right)\hat j\\
\left| {\vec a} \right| = 50
\end{array}$
$\begin{array}{l}
\vec r = \left( {15{t^2}} \right)\hat i + \left( {4 - 20{t^2}} \right)\hat j\\
\vec v = \frac{{d\vec r}}{{dt}} = \left( {30t} \right)\hat i - \left( {40t} \right)\hat j\\
\vec a = \frac{{d\vec v}}{{dt}} = \left( {30} \right)\hat i - \left( {40} \right)\hat j\\
\left| {\vec a} \right| = 50
\end{array}$
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