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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) \( 3 \alpha+7 \beta \) |
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Answer» \(v = \alpha t + \beta t ^2\) \(\because v= \frac{dx}{dt}\) \(\frac{dx}{dt}= \alpha t + \beta t ^2\) \(\int dx = \int\limits_1^2 (\alpha t + \beta t^2)dt\) \(= \left[\frac{\alpha t^2}2+\frac{\beta^3}3\right]_1^2\) \(= \left[2\alpha + \frac {8\beta}3 - \frac \alpha 2 - \frac\beta 3\right]\) \(x = \left[ \frac{3\alpha}2 + \frac{7\beta}3\right]\) |
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