Question

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 1

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 ,

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

Answer 2

Explanation:

$\Large\boxed{\sf{(b)\quad\!18\ m\ s^{-1}}}$

Given, acceleration as a function in time (in $\sf{m\ s^{-2}}$

. Then,

$\longrightarrow\sf{\dfrac{dv}{dt}=3t^2+2t+2}$

$\longrightarrow\sf{dv=(3t^2+2t+2)\ dt}⟶dv=(3t 2)$

Now we have to integrate both sides.

Integration should be done for a time from t = 0 seconds to t = 2 seconds.

Integration should be done for a velocity from v = u = 2 $\sf{m\ s^{-1}}m s$

• where v is the velocity at the time t = 2 seconds.

So,

$\displaystyle\longrightarrow\sf{\int\limits_2^vdv=\int\limits_0^2(3t^2+2t+2)\ dt}$

$\displaystyle\longrightarrow\sf{\big[v\big]_2^v=\int\limits_0^23t^2\ dt+\int\limits_0^22t\ dt+\int\limits_0^22\ dt}⟶[v]$

$\displaystyle\longrightarrow\sf{v-2=3\int\limits_0^2t^2\ dt+2\int\limits_0^2t\ dt+2\int\limits_0^2\ dt}$

$\displaystyle\longrightarrow\sf{v-2=3\left[\dfrac{t^3}{3}\right]_0^2+2\left[\dfrac{t^2}{2}\right]_0^2+2\big[t\big]_0^2}$

$\displaystyle\longrightarrow\sf{v-2=3\cdot\dfrac{2^3-0^3}{3}+2\cdot\dfrac{2^2-0^2}{2}+2(2-0)}$

$\displaystyle\longrightarrow\sf{v-2=8-0+4-0+2\times2}$

$\displaystyle\longrightarrow\sf{v-2=8+4+4}$

$\displaystyle\longrightarrow\sf{v-2=16}⟶v−2=16$

$\tt ⟶ v=18 m s −1$