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

(B) $\frac{3}{2} \alpha+\frac{7}{3} \beta$

(A) $\frac{\alpha}{2}+\frac{\beta}{3}$

(C) $\frac{3}{2} \alpha+\frac{7}{2} \beta$

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Kinematics

The instantaneous velocity of a particle moving in a straight line is given as

v = α t + β t^{2}

, where

α

and

β

are constants. The distance travelled by the particle between 1 s and 2 s is :

(A) $\frac{α}{2} + \frac{β}{3}$
(B) $\frac{3}{2} α + \frac{7}{3} β$ ✓ Correct
(C) $\frac{3}{2} α + \frac{7}{2} β$
(D) $3 α + 7 β$

Answer: (B)

Solution

Distance is

s = \int_{1}^{2} v d t = \int_{1}^{2} (α t + β t^{2}) d t = [\frac{α t ^{2}}{2} + \frac{β t ^{3}}{3}]_{1}^{2} = (2 α + \frac{8 β}{3}) - (\frac{α}{2} + \frac{β}{3}) = \frac{3 α}{2} + \frac{7 β}{3}

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Single correct option

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