The instantaneous velocity of a particle moving in a straight line is given as V = α t + β t², where α and β are constants. The distance travelled by the particle between 1s and 2s is: (a) 3 α + 7 β (b) 3/2α + 7/3β (c) (α)/(2) + (β)/(3) (d) 3/2α + 7/2β

Answer: (b). Distance = ∫₁² (α t + β t²) dt = (α)/(2)(2² - 1²) + (β)/(3)(2³ - 1³) = 3/2α + 7/3β So the answer is (b).

The instantaneous velocity of a particle moving in a straight line is… | JEE Main 2021 (25 Jul, Shift 2)

Q24JEE Main 2021 · 25 JUL · SHIFT 2Easy

Velocity, Acceleration & Variable Motion

The instantaneous velocity of a particle moving in a straight line is given as $V = \alpha t + \beta t^2$ , where $\alpha$ and $\beta$ are constants. The distance travelled by the particle between 1s and 2s is:

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Answer: (b)

Distance $= \int_1^2 (\alpha t + \beta t^2)\,dt$
$= \dfrac{\alpha}{2}(2^2 - 1^2) + \dfrac{\beta}{3}(2^3 - 1^3)$
$= \dfrac{3}{2}\alpha + \dfrac{7}{3}\beta$
So the answer is (b).

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Velocity, Acceleration & Variable Motion question from JEE Main 2021 (25 Jul, Shift 2)

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