33. If the velocity of a particle is v = At + Bt2, where A and B are constants, then the distance travelled by it between 1 s and 2 s is:

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

(a)  \(\frac{3}{2}\) A + 4 B

(b)    3 A + 7 B

(c)  \(\frac{3}{2}\) A + \(\frac{7}{3}\) B

(d)  \(\frac{A}{2}\) + \(\frac{B}{3}\)

Answer :  (c)  \(\frac{3}{2}\) A + \(\frac{7}{3}\) B

\( v = At + Bt^2 \)

a = \( \frac{dv}{dt} \) = A + 2Bt

Acceleration is changing, so equations of motion cannot be used.

\( a = f(t) \)

\( \frac{dv}{dt} \) = f(t)

\( \int dv = \int f(t)\, dt \)

\( v = At + Bt^2 \)

\( \frac{dx}{dt} \) = At + Bt^2

\( \int dx = \int (At + Bt^2)\, dt \)

= [ \( \frac{A t^2}{2} \) + \( \frac{B t^3}{3} \) ]\(_{t=1}^{t=2}\)

= [ \( \frac{A \, 2^2}{2} \) + \( \frac{B \, 2^3}{3} \) ] − [ \( \frac{A \, 1^2}{2} \) + \( \frac{B \, 1^3}{3} \) ]

\( \rightarrow x_2 - x_1 \) = \( \frac{3A}{2} \) + \( \frac{7B}{3} \)