Question

The equation of a displacement-time curve of a particle is given by X =10t- 5t + 6 = 0. Find the instantaneous velocities at t = 2sec and t = 5sec.
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Answer 1

$\\ \rm\hookrightarrow v={\displaystyle{\int}}sdt$

Now

$\\ \rm\hookrightarrow v={\displaystyle{\int}^5_2}10t-5t+6$

$\\ \rm\hookrightarrow v={\displaystyle{\int}^5_2}5t+6$

$\boxed{\sf {\displaystyle{\int}}x^n dx=\dfrac{x^{n+1}}{n+1}}$

$\\ \rm\hookrightarrow v=\left[\dfrac{5t^2}{2}+6t\right]^5_2$

$\\ \rm\hookrightarrow v=\dfrac{5\left\{5)^2-(2)^2\right\}}{2}+6(5-2)$

$\\ \rm\hookrightarrow v=\dfrac{5(25-4)}{2}+6(3)$

$\\ \rm\hookrightarrow v=\dfrac{5(21)}{2}+18$

$\\ \rm\hookrightarrow v=\dfrac{105}{2}+18$

$\\ \rm\hookrightarrow v=52.5+18$

$\\ \rm\hookrightarrow v=70.5m/s$