ExamBro
ExamBro
JEE Mains · Physics · STD 11 - 3.2 motion in plane

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 \(t = 1\) ?

  1. A \(40\)
  2. B \(100\)
  3. C \(25\)
  4. D \(50\)
Verified Solution

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\\…

Same subject
Explore more questions on app