Question

The acceleration of a particle is increasing linearly with time $t$ as $bt$. The particle starts from the origin with an initial velocity $v_0$. The distance travelled by the particle in time $t$ will be

• A. $v_{0}t+\frac{1}{3}b{t}^2$
• B. $v_{0}t+\frac{1}{3}b{t}^3$
• C. $v_{0}t+\frac{1}{6}b{t}^3$
• D. $v_{0}t+\frac{1}{2}b{t}^2$

Answer 1

Answer: (C)

$v_{0}t+\frac{1}{6}b{t}^3$

here it is given that $a=bt$
wkt, $a=\frac{changeinvelocity}{changeintime}=\frac{dv}{dt}$
$\therefore \frac{dv}{dt}=bt$ now integrating this we will get $v$
$v=\frac{b{t}^2}{2}+c$
At $t=0s,v=v_0,Thus{v}_0=c$
$\therefore v=\frac{b{t}^2}{2}+v_0$
Now. $v=\frac{changeinposition}{changeintime}=\frac{ds}{dt}=\frac{b{t}^2}{2}+v_0$
$\therefore ds=(\frac{b{t}^2}{2}+v_0)dt$
$\therefore$ Integrating the above we get, $s=\frac{b{t}^3}{6}+v_{0}t+c^{{}^{\prime }}att=0,s=0\therefore c{}^{\prime }=0$
i.e. $s=\frac{b{t}^3}{6}+v_{0}t$