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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 \(....\,ms ^{-1}\)

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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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