The acceleration of a particle is increasing linearly with time $t$ as $bt$. The particle starts from the orig

Question

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

Accepted Answer

1. **Identify the Relationship:** Acceleration $a$ is given as a function of time: $a = bt$. Since acceleration is non-uniform, we must use calculus (integration) to find velocity and position .
2. **Find Velocity ($v$):** Acceleration is the derivative of velocity ($a = \frac{dv}{dt}$). Integrating acceleration with respect to time:
   $$ \int_{v_0}^{v} dv = \int_{0}^{t} a \, dt = \int_{0}^{t} bt \, dt $$
   $$ v - v_0 = \left[ \frac{bt^2}{2} \right]_0^t \implies v = v_0 + \frac{1}{2}bt^2 $$
3. **Find Distance ($x$):** Velocity is the derivative of position ($v = \frac{dx}{dt}$). Integrating velocity with respect to time:
   $$ \int_{0}^{x} dx = \int_{0}^{t} v \, dt = \int_{0}^{t} \left( v_0 + \frac{1}{2}bt^2 \right) dt $$
   $$ x = \left[ v_0t + \frac{1}{2}b \cdot \frac{t^3}{3} \right]_0^t $$
   $$ x = v_0t + \frac{1}{6}bt^3 $$

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