Question

The acceleration of a particle is increasing linearly with time t as bt . The particle starts from the origin with an initial velocity $v_o$. 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

The correct option is C $v_{0}t+\frac{1}{6}b{t}^3$

$\frac{dv}{dt}=bt\Rightarrow dv=btdt\Rightarrow v=\frac{b{t}^2}{2}+K_1$

At $t=0,v=v_o\Rightarrow K_1=v_0$

We get$v=\frac{1}{2}b{t}^2+v_0$

Again $\frac{dx}{dt}=\frac{1}{2}b{t}^2+v_0\Rightarrow x=\frac{1}{2}\frac{b{t}^2}{3}+v_{0}t+K_2$

At $t=0,x=0\Rightarrow K_2=0$

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