Question

the acceleration of particle is given by an equation a=3t²+2t+12 if initial velocity of particle is 5m/s then find the velocity at end of 2 sec​

Answer 1

Given :

Acceleration of particle is given by

• a = 3t² + 2t + 12

Initial velocity of particle = 5m/s

To Find :

Velocity of body at the end of 2s.

Solution :

❖ In order to find velocity of particle, we have to integrate the given acceleration equation with respect to time.

$\displaystyle\dag\:\underline{\boxed{\bf{\orange{\int dv=\int a \cdot dt}}}}$

$\displaystyle\sf:\implies\:\int\limits_{u}^{v} dv=\int\limits_{0}^{2} (3t^2+2t+12)\cdot dt$

$\sf:\implies\:\left[v\right]_u^v=\left[\dfrac{3t^3}{3}+\dfrac{2t^2}{2}+12t\right]_0^2$

$\sf:\implies\:v-u=t^3+t^2+12t$

$\sf:\implies\:v-5=(2)^3+(2)^2+12(2)$

$\sf:\implies\:v-5=8+4+24$

$\sf:\implies\:v-5=36$

$:\implies\:\underline{\boxed{\bf{\gray{v=41\:ms^{-1}}}}}$

Answer 2

Given :

Acceleration of particle is given by

a = 3t² + 2t + 12

Initial velocity of particle = 5m/s

To Find :

Velocity of body at the end of 2s.

Solution :

❖ In order to find velocity of particle, we have to integrate the given acceleration equation with respect to time.

$\displaystyle\dag\:\underline{\boxed{\bf{\orange{\int dv=\int a \cdot dt}}}}$

$\displaystyle\sf:\implies\:\int\limits_{u}^{v} dv=\int\limits_{0}^{2} (3t^2+2t+12)\cdot dt$

$\sf:\implies\:\left[v\right]_u^v=\left[\dfrac{3t^3}{3}+\dfrac{2t^2}{2}+12t\right]_0^2$

$\sf:\implies\:v-u=t^3+t^2+12t$

$\sf:\implies\:v-5=(2)^3+(2)^2+12(2)$

$\sf:\implies\:v-5=8+4+24$

$\sf:\implies\:v-5=36$

$:\implies\:\underline{\boxed{\bf{\gray{v=41\:ms^{-1}}}}}$