Kinematics

JEE Main level, single correct, medium

136 of 308

Question 136 of 308

The velocity of a particle is v=v0+gt+ft2\mathrm{v}=\mathrm{v}_0+\mathrm{gt}+\mathrm{ft}^2. If its position is x=0\mathrm{x}=0 at t=0\mathrm{t}=0, then its displacement after unit time (t=1)(t=1) is

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Answer: (B) v0+g/2+f/3\mathrm{v_0+g / 2+f / 3}

∫0xdx=∫01(V0+gt+ft2)dt\mathrm{\int_0^x d x=\int_0^1\left(V_0+g t+\mathrm{ft}^2\right) d t}
x=v0+g(12)+f(13)\mathrm{x=v_0+g\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)}

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