The position vector of a particle changes with time according to the relation \(\vec r(t)=15t^2\hat i+(4-20t^2)\hat j\). What is the magnitude of the acceleration at \(t=1\) ?
Show Hint
- 50
- 25
- 40
- 100
The Correct Option is A
Solution and Explanation
\(\vec a=\frac{d\vec v}{dt}=30\,\hat i-40\,\hat j\) (constant, independent of \(t\)).
\(|\vec a|=\sqrt{30^2+40^2}=\sqrt{2500}=50\).
Answer: (1) 50
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