Question

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 \( 1 s \) and \( 2 s \) is

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

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

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

(4) \( 3 \alpha+7 \beta \)

Answer 1

\(v = \alpha t + \beta t ^2\)

\(\because v= \frac{dx}{dt}\)

\(\frac{dx}{dt}= \alpha t + \beta t ^2\)

\(\int dx = \int\limits_1^2 (\alpha t + \beta t^2)dt\)

\(= \left[\frac{\alpha t^2}2+\frac{\beta^3}3\right]_1^2\)

\(= \left[2\alpha + \frac {8\beta}3 - \frac \alpha 2 - \frac\beta 3\right]\)

\(x = \left[ \frac{3\alpha}2 + \frac{7\beta}3\right]\)