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/s
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