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The position vector of a particle changes with time according to the relation \(\vec r\left( {\rm{t}} \right) = 15{{\rm{t}}^2}{\rm{\hat i}} + \left( {4 - 20{{\rm{t}}^2}} \right){\rm{\hat j}}\). What is the magnitude of the acceleration at t = 1?
1. 40
2. 25
3. 100
4. 50

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Correct Answer - Option 4 : 50

Concept:

Given the position vector of the particle as a function of time t,

Then,

Velocity:

The rate of change of displacement of an object (displacement over elapsed time) is velocity.

Velocity is given by \(v = \frac{{dx}}{{dt}}\)

Acceleration:

The rate of change of velocity is acceleration. Like velocity, acceleration is a vector and has both magnitude and direction. Acceleration is given by \(a = \frac{{dv}}{{dt}}\)

Calculation:

Given,

Time t = 1 sec

Position vector is given by,

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

Velocity is termed as the change in their displacement.

\(\vec v = \frac{{d\vec r}}{{dt}}\)

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

\(\Rightarrow \vec v = 30t\hat i - 40t\hat j\)

Acceleration is termed as the change in their velocity.

\(\vec a = \frac{{d\vec v}}{{dt}}\)

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

\(\vec a = 30\hat i - 40\hat j\)

\(\therefore \left| {\vec a} \right| = \sqrt {{{30}^2} + {{40}^2}} = 50\;{\rm{m}}/{{\rm{s}}^2}\)

Therefore, the magnitude of the acceleration at t = 1 is 50 m/s2

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