Kinematics

JEE Main level, single correct, medium

29 of 308

Question 29 of 308

The instantaneous velocity of a particle moving in a straight line is given as v=αt+βt2v=\alpha t+\beta {t}^{2}, where α\alpha and β\beta are constants. The distance travelled by the particle between 1s1s and 2s2s is:

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Answer: (B) 32α+73β\dfrac{3}{2}\alpha +\dfrac{7}{3}\beta

V=αt+βt2V=\alpha t+\beta {t}^{2}
dsdt=αt+βt2\dfrac{ds}{dt}=\alpha t+\beta {t}^{2}
∫S1S2ds=∫12(αt+βt2)dt{\int }_{{S}_{1}}^{{S}_{2}}ds={\int }_{1}^{2}\left(\alpha t+\beta {t}^{2}\right)dt
S2−S1=[αt22+βt33]12{S}_{2}-{S}_{1}={\left[\dfrac{\alpha {t}^{2}}{2}+\dfrac{\beta {t}^{3}}{3}\right]}_{1}^{2}
As the particle is not changing direction So distance == displacement. Distance =[α[4−1]2+β[8−1]3]=\left[\dfrac{\alpha [4-1]}{2}+\dfrac{\beta [8-1]}{3}\right]
=3α2+7β3=\dfrac{3\alpha }{2}+\dfrac{7\beta }{3}

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