Question

The acceleration of a particle is increasing linearly with time $t$ as $\beta t$. If the particle starts from origin with initial velocity $u$, the distance travelled by it in $t$ second is

• A. $ut+\beta t^3$
• B. $ut+\frac{1}{2}\beta t^3$
• C. $ut+\frac{1}{3}\beta t^3$
• D. $ut+\frac{\beta t^3}{6}$

Answer 1

Answer: (D)

$ut+\frac{\beta t^3}{6}$

$a=\beta t\Rightarrow \frac{dv}{dt}=\beta t\Rightarrow dv=\beta tdt$

Integrating both sides with given limits, we get
$v-u=\beta \frac{t^2}{2}$

Now,
$v=\frac{dx}{dt}=u+\beta \frac{t^2}{2}$
$dx=udt+\beta \frac{t^2}{2}dt$

Integrating both sides with given limits, we get
$\Delta x=ut+\frac{\beta t^3}{6}$