JEE Mains · Physics · STD 11 - Motion in straight line
- JEE Mains 2023
- Easy
- Numerical
The distance travelled by an object in time \(t\) is given by \(s =(2 \cdot 5) t ^2\). The instantaneous speed of the object at \(t =5\,s\) will be \(....\,ms ^{-1}\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(25\)
Step-by-step Solution
Detailed explanation
Distance \((s)=(2.5) t^2\)
Speed \(( v )=\frac{ ds }{ dt }=\frac{ d }{ dt }\left\{(2.5) t ^2\right\}\)
\(v =5\,t\)
At \(t =5, v =5 \times 5=25\,m / s\)
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