Question

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

• A. $v_0t+\frac{1}{3}bt^2$
• B. $v_0 t+\frac{1}{2}bt^2$
• C. $v_0 t+\frac{1}{6}bt^3$
• D. $v_0 t+\frac{1}{3}bt^3$

Answer 1

The Correct Option is C

To find the distance travelled by the particle in time t, given that the acceleration is increasing linearly with time as bt and the initial velocity is v_0, we need to perform the following steps:

1. The acceleration a of the particle is given as a = bt. This is a function of time.
2. First, we need to find the velocity. Since acceleration is the derivative of velocity with respect to time, we integrate the acceleration function with respect to time to get velocity:
   v = \int a \, dt = \int bt \, dt = \frac{1}{2}bt^2 + C_1

   where C_1 is the constant of integration.

3. We know the initial condition: at time t = 0, the velocity v = v_0. Substituting these values will help us determine C_1:
   v_0 = \frac{1}{2}b(0)^2 + C_1 \Rightarrow C_1 = v_0

   Thus, the velocity as a function of time becomes:

   v = \frac{1}{2}bt^2 + v_0
4. Next, we find the distance travelled by integrating the velocity function. Distance s is the integral of velocity with respect to time:
   s = \int v \, dt = \int \left(\frac{1}{2}bt^2 + v_0\right) \, dt = \frac{1}{6}bt^3 + v_0 t + C_2

   where C_2 is another constant of integration.

5. Using the initial condition: at t = 0, s = 0, let's find C_2:
   0 = \frac{1}{6}b(0)^3 + v_0(0) + C_2 \Rightarrow C_2 = 0

   Thus, the distance travelled by the particle at time t is:

   s = v_0 t + \frac{1}{6}bt^3

Therefore, the correct answer is v_0 t + \frac{1}{6}bt^3.