Question

The acceleration of a particle in m/s2 is given by a = 4t 2+2t+2, where time t is in second. If the particle starts with a velocity v = 2 m/s at t=0, find the velocity at t = 5 s.

Answer 1

Answer:

• The acceleration of a particle in m/s2 is given by a = 4t²+2t+2, where time t is in second. The particle starts with a velocity v = 2 m/s at t=0.

• Acceleration:

$\implies \sf a = 4t^2 + 2t + 2$

$\implies \sf a = \dfrac{dv}{dt}$

$\implies \sf dv = a.dt$

•Integrating both the sides:

$\displaystyle\implies \sf \int\limits_{u}^{v}dv = \int\limits_{0}^{t}a.dt$

$\displaystyle\implies \sf v - u = \int\limits_{0}^{t = 5}( {4t}^{2} + 2t + 2).dt$

$\displaystyle\implies \sf v - 2 = \int\limits_{0}^{t = 5}( {4t}^{2} + 2t + 2).dt$

$\displaystyle\implies \sf v - 2 = \left. \bigg\{\dfrac{ {4t}^{3} }{3} + \dfrac{ {2t}^{2} }{2} + 2t\bigg\}\right| _{0}^{5}$

$\displaystyle\implies \sf v - 2 = \left. \bigg\{\dfrac{ {4t}^{3} }{3} + {t}^{2} + 2t\bigg\}\right| _{0}^{5}$

$\displaystyle\implies \sf v - 2 = \bigg\{ \bigg(\dfrac{ {4(5)}^{3} }{3} - \dfrac{ {4(0)}^{3} }{3} \bigg) + ( {5}^{2} - {0}^{2} ) + (2 \times 5 - 2 \times 0)\bigg \}$

$\displaystyle\implies \sf v - 2 = \dfrac{ 500 }{3} + 25 + 10$

$\displaystyle\implies \sf v = \dfrac{ 500 }{3} + 25 + 10 + 2$

$\displaystyle\implies \sf v = \dfrac{ 500 }{3} +37$

$\displaystyle\implies \sf v = \dfrac{ 500 + 111}{3}$

$\displaystyle\implies \sf v = \dfrac{ 611}{3}$

$\displaystyle\implies \underline{ \boxed{ \orange{ \bf v = 203.67 \: m/s}}}$