JEE Mains · Physics · STD 11 - Motion in straight line
- JEE Mains 2021
- Easy
- MCQ
The velocity of a particle is \(v = v _{0}+ gt + Ft ^{2}\). Its position is \(x=0\) at \(t=0\); then its displacement after time \(( t =1)\) is :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(v_{0}+\frac{g}{2}+\frac{F}{3}\)
Step-by-step Solution
Detailed explanation
\(v = v _{0}+ g t + Ft ^{2}\)
\(\frac{ ds }{ dt }= v _{0}+ gt + Ft ^{2}\)
\(\int ds =\int_{0}^{1}\left( v _{0}+ gt + Ft ^{2}\right) dt\)
\(s=\left[v_{0} t+\frac{g t^{2}}{2}+\frac{F t^{3}}{3}\right]_{0}^{1}\)
\(s=v_{0}+\frac{g}{2}+\frac{F}{3}\)
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