Question

The acceleration of a particle b increasing linearly with time t as bt. The particle starts from origin with an initial velocity $V_0$. The distance traveled by the particle in time t will be______

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

Answer 1

Answer: (C)

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

Acceleration$=\frac{d^{2}s}{d{t}^2}=bt$

Integrating above equation

$V=\frac{ds}{dt}=\frac{b{t}^2}{2}+c$

at $t=0,V_0=u$

So, $u=0+c$

$\frac{ds}{dt}=\frac{b{t}^2}{2}+u$

Again intrigate $\therefore S=\frac{b{t}^3}{6}+ut+d$

at $t=0,s=0,$, So,$d=0$

$S=\frac{b}{6}{t}^3+ut$

$u=V_0$

$\therefore S=V_{0}t+\frac{1}{6}b{t}^3$