Question

(d) 0.5
The acceleration 'a' in m/s- of a particle is given
by a = 3t? + 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 is:
(a) 12 m/s
(b) 18 m/s
(c) 27 m/s
(d) 36 ​

Answer 1

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

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

$\longrightarrow\sf{a=3t^2+2t+2}$

We know that $\sf{a=\dfrac{dv}{dt}.}$ Then,

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

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

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 $\sf{m\ s^{-1}}$ to v = v (Say)

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

So,

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

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

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

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

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

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

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

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

$\displaystyle\longrightarrow\sf{\underline{\underline{v=18\ m\ s^{-1}}}}$