Distance travelled by an object in time t is given by s =(2 · 5) t ^2 . instantaneous speed of obje

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2.Motion in Straight Line

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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

$25$

(JEE MAIN-2023)

Solution

Distance $(s)=(2.5) t^2$

Speed $( v )=\frac{ ds }{ dt }=\frac{ d }{ dt }\left\{(2.5) t ^2\right\}$

$v =5\,t$

At $t =5, v =5 \times 5=25\,m / s$

Standard 11

Physics

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