The acceleration 'a' in m/s^2 of a particle is given by a= 3t^2+2t+2 w

1.	The acceleration 'a' in m/s^2 of a particle is given by a= 3t^2+2t+2 where t is the time. If the particle starts out with a velocity u = 2m/s at t = 0, then the velocity at the end of 2 second will be ?
Answer» NEED to FinD :- The velocity at the end of 2 SECONDS . $\red{\frak{Given}}\begin{cases}\sf The \ accl^n \ of \ particle \ is \ a \ = \ 3t^2+2t + 2 .\\\textsf{ The particle starts out with a velocity u = 2m/s at t = 0 } \end{cases}$ We know the acceleration is the rate of change of velocity . So , that , $\sf\dashrightarrow \dfrac{dv}{dt}= a$ • Now transposing dt to RHS , we have , $\sf\dashrightarrow dv = a . dt \\\\\sf\dashrightarrow \displaystyle \sf\int dv = \int a . dt \\\\\sf\dashrightarrow \displaystyle\sf \int (3t^2+2t+2) .dt = \int dv$ LET the velocity at the end of 2 s be v . Therefore the upper limit of dv will be v and lower limit will be 2 . Also TIME will be integrated from t = 0 to t = 2 s . Putting the limits , we have , $\dashrightarrow \displaystyle\sf \int_0 ^2 (3t^2+2t+2) .dt = \int_2^v dv \\\\\sf\dashrightarrow \displaystyle\sf (v-2) = \bigg[ \dfrac{3t^{2+1}}{2+1} + \dfrac{2t^{1+1}}{1+1} + 2t \bigg] ^2_0 \\\\\dashrightarrow \displaystyle \sf (v - 2) = \bigg[ \dfrac{3t^3}{3}+\dfrac{2t^2}{2}+ 2t\bigg]_0^2 \\\\\dashrightarrow \sf (v-2) = [ t^3 + t^2 + 2t]^2_0 \\\\\sf\dashrightarrow (v -2) = 2^3 + 2^2 +2(2) - 0 \\\\\sf\dashrightarrow (v-2) = 8 + 4 + 4 - 0 \\\\\sf\dashrightarrow (v - 2) = 16 \\\\\sf\dashrightarrow v = ( 16 + 2) m/s \\\\\sf\dashrightarrow \underset{\blue{\sf Required\ velocity}}{\underbrace{\boxed{\pink{\frak{ Velocity_{(At\ end\ of \ 2s)}= 18m/s }}}}}$

The acceleration 'a' in m/s^2 of a particle is given by a= 3t^2+2t+2 where t is the time. If the particle starts out with a velocity u = 2m/s at t = 0, then the velocity at the end of 2 second will be ?

Answer»

NEED to FinD :-

The velocity at the end of 2 SECONDS .

$\red{\frak{Given}}\begin{cases}\sf The \ accl^n \ of \ particle \ is \ a \ = \ 3t^2+2t + 2 .\\\textsf{ The particle starts out with a velocity u = 2m/s at t = 0 } \end{cases}$

We know the acceleration is the rate of change of velocity . So , that ,

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

• Now transposing dt to RHS , we have ,

$\sf\dashrightarrow dv = a . dt \\\\\sf\dashrightarrow \displaystyle \sf\int dv = \int a . dt \\\\\sf\dashrightarrow \displaystyle\sf \int (3t^2+2t+2) .dt = \int dv$

LET the velocity at the end of 2 s be v . Therefore the upper limit of dv will be v and lower limit will be 2 . Also TIME will be integrated from t = 0 to t = 2 s . Putting the limits , we have ,

$\dashrightarrow \displaystyle\sf \int_0 ^2 (3t^2+2t+2) .dt = \int_2^v dv \\\\\sf\dashrightarrow \displaystyle\sf (v-2) = \bigg[ \dfrac{3t^{2+1}}{2+1} + \dfrac{2t^{1+1}}{1+1} + 2t \bigg] ^2_0 \\\\\dashrightarrow \displaystyle \sf (v - 2) = \bigg[ \dfrac{3t^3}{3}+\dfrac{2t^2}{2}+ 2t\bigg]_0^2 \\\\\dashrightarrow \sf (v-2) = [ t^3 + t^2 + 2t]^2_0 \\\\\sf\dashrightarrow (v -2) = 2^3 + 2^2 +2(2) - 0 \\\\\sf\dashrightarrow (v-2) = 8 + 4 + 4 - 0 \\\\\sf\dashrightarrow (v - 2) = 16 \\\\\sf\dashrightarrow v = ( 16 + 2) m/s \\\\\sf\dashrightarrow \underset{\blue{\sf Required\ velocity}}{\underbrace{\boxed{\pink{\frak{ Velocity_{(At\ end\ of \ 2s)}= 18m/s }}}}}$