Question

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

Answer 1

Answer on Question #40689, Physics, Mechanics

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

Solution:

Given:

$$a(t) = bt$$

The derivative of a distance function represents instantaneous velocity and that the derivative of the velocity function represents instantaneous acceleration at a particular time.

Thus,

$$a = \frac{dv}{dt} = bt$$

Integrating, the equation

$$\int dv = \int bt \, dt$$ $$v = \frac{bt^2}{2} + C$$

where $C$ is constant of integration.

At $t = 0$, $v = v_0$, hence $C = v_0$ and equation becomes

$$v = \frac{bt^2}{2} + v_0$$

We have that

$$v = \frac{dx}{dt}$$

where $x$ is distance

$$\frac{dx}{dt} = \frac{bt^2}{2} + v_0$$

Integrating

$$\int dx = \int \left(\frac{bt^2}{2} + v_0\right) dt$$ $$x = \frac{bt^3}{2} + v_0 t + x_0$$

where $x_0$ is constant of integration.

At $t = 0$, $x = 0$, hence $x_0 = 0$ and equation becomes

$$x = \frac{bt^3}{6} + v_0 t$$

Answer. $x = \frac{bt^3}{6} + v_0 t$.