If the velocity of a particle is υ = At + Bt^ 2 , where A and B are constants, then the distance travelled by it between 1s and 2s is:

Correct answer: C. (3)/(2)A + (7)/(3)B V = At + B t ^ 2 X = ∫ _ 0 ^ t Vdt ⇒ X = A t ^ 2 2 + B t ^ 3 3 t = 1 X _ 1 = (A)/(2) + (B)/(3) t = 2 X _ 2 = 2A + (8B)/(3) X _ 2 - X _ 1 = (3A)/(2) + (7B)/(3)

NEET 2016

Exam: NEET
Year: 2016
Subject: Physics
Chapter: Motion in One Dimension

NEET 2016 · Physics · Motion in One Dimension

If the velocity of a particle is υ = At + Bt^ 2 , where A and B are...

If the velocity of a particle is

υ = At + Bt^{2}

, where

\text{A}

and

\text{B}

are constants, then the distance travelled by it between

\text{1}\text{s}

and

\text{2}\text{s}

is:

Options

A $\dfrac{3}{2}A + 4B$
B $3A + 7B$
C $\dfrac{3}{2}A + \dfrac{7}{3}B$
D $\dfrac{A}{2} + \dfrac{B}{3}$

Correct answer

C. $\dfrac{3}{2}A + \dfrac{7}{3}B$

Step-by-step solution

V = At + B{t}^{2}

X = {∫}_{0}^{t} Vdt ⇒

X = \dfrac{A{t}^{2}}{2} + \dfrac{B{t}^{3}}{3}

t = 1

{X}_{1} = \dfrac{A}{2} + \dfrac{B}{3}

t = 2

{X}_{2} = 2A + \dfrac{8B}{3}

{X}_{2} - {X}_{1} = \dfrac{3A}{2} + \dfrac{7B}{3}

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