JEE Mains · Physics · STD 11 - Motion in straight line
- JEE Mains 2023
- Easy
- Numerical
The position of a particle related to time is given by \(x =\left(5 t ^2-4 t +5\right) m\). The magnitude of velocity of the particle at \(t=2\,s\) will be \(...........ms^{-1}\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(16\)
Step-by-step Solution
Detailed explanation
\(x=5 t^2-4 t+5\)
\(v=10 t-4\)
\(\text { At } t=2 s \quad v=16\,m / s\)
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