Question

The acceleration ‘a’ (in m/s2) of a particle is given by a = 3t2 + 2t – 2, where t is time in second. If the particle starts with 10 m/s at t = 0, then the velocity at t = 2 s is

Answer 1

$\Large{\bf{\underline{\underline{\green{Solution}}}}}$

$\sf{\green{Given}}\begin{cases}</p><p>\sf{a = 3t^{2} + 2t - 2}\\ \\</p><p>\sf{Initial \: velocity \: = 10\:m/s}\\ \\</p><p>\sf{Initial \: time \: = 0\: s}\\ \\</p><p>\sf{Final \: time \: = 2\: s}\\</p><p>\end{cases}$

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$</p><p>\large{\sf{\star\: \: a = 3t^{2} + 2t - 2}}\\</p><p>$

$</p><p>\sf{\implies\: \frac{dv}{dt} = 3t^{2} + 2t - 2}\\</p><p>$

$</p><p>\sf{\purple{Using\:integration}}\\</p><p>$

$</p><p>\sf{\implies\: \int dv = \int (3t^{2} + 2t - 2)dt}\\</p><p>$

Let the initial velocity be 'u' and final velocity be 'v' respectively.

And the initial time and final time be 't1' and 't2' respectively

$</p><p>\sf{\implies\: \int \limits_{u}^{v} dv = \int \limits_{t_{1}}^{t_{2}} (3t^{2} + 2t - 2)dt}\\</p><p>$

$</p><p>\sf{\implies\: \int \limits_{10}^{v} dv = \int \limits_{0}^{2} (3t^{2} + 2t - 2)dt}\\</p><p>$

$</p><p>\sf{\implies\: \left[ v - 10 \right] = \left[ \frac{3t^{3}}{3}+ \frac{2t^{2}}{2} - 2t \right] }\\</p><p>$

Putting t = 2

$</p><p>\sf{\implies\: v - 10 = (2)^{3} + (2)^{2} - 2(2)}\\</p><p>$

$</p><p>\sf{\implies\: v - 10 = 8 + \cancel{4} - \cancel{4}}\\</p><p>$

$</p><p>\Large{\boxed{\sf{\green{\therefore\: v = 18 \: m/s}}}}\\</p><p>$

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Note:

$</p><p>\sf{\implies\: \int \limits_{2}^{3} v^{2} = \left[ \frac{(3)^{3}}{3} - \frac{(2)^{3}}{3} \right] }\\</p><p>$