JEE Mains · Physics · STD 11 - 2. motion in straight line
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 \(....\,ms ^{-1}\)
- A \(12.5\)
- B \(62.5\)
- C \(5\)
- D \(25\)
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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