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

Answer 1

Answer: (C)

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

Answer 2

$\int_{v_{1}}^{v_{2}}{}$dv = $\int_{t_{1}}^{t_{2}}{adt}$ =$\int_{t_{1}}^{t_{2}}{(bt)dt}$

$\Rightarrow v_{2} - v_{1} = \left( \frac{bt^{2}}{2} \right)_{t_{1}}^{t_{2}}$

⇒ $v_{2} = v_{1} + \left( \frac{bt^{2}}{2} \right)_{0}^{t} = v_{0} + \frac{bt^{2}}{2}$

⇒ $S = \int_{}^{}v_{0}dt + \int_{}^{}{\frac{bt^{2}}{2}dt} = v_{0}t + \frac{1}{6}bt^{3}$