Question

A particle moves on x-axis such that its KE varies as a relation KE = 3t² then the average kinetic (C) of a particle in 0 to 2 sec is given by:

• A. 6 joule
• B. 8 joule
• C. 4 joule
• D. 16 joule

Answer 1

Answer: (C)

4 joule

Answer 2

The kinetic energy of the particle is given by the relation:

$KE = 3t^2$

We need to find the average kinetic energy of the particle in the time interval from $t=0$ to $t=2$ seconds.

The average value of a function $f(t)$ over an interval $[a, b]$ is given by the formula:

$f_{avg} = \frac{1}{b-a} \int_{a}^{b} f(t) dt$

In this case, $f(t) = KE(t) = 3t^2$, $a=0$, and $b=2$.

Substitute these values into the formula:

$KE_{avg} = \frac{1}{2-0} \int_{0}^{2} 3t^2 dt$ $KE_{avg} = \frac{1}{2} \int_{0}^{2} 3t^2 dt$

Now, evaluate the definite integral:

$\int 3t^2 dt = 3 \frac{t^{2+1}}{2+1} = 3 \frac{t^3}{3} = t^3$

Apply the limits of integration:

$\int_{0}^{2} 3t^2 dt = [t^3]_{0}^{2} = (2)^3 - (0)^3 = 8 - 0 = 8$

Now substitute the result of the integral back into the average kinetic energy equation:

$KE_{avg} = \frac{1}{2} (8)$ $KE_{avg} = 4 \text{ joule}$

The average kinetic energy of the particle from 0 to 2 seconds is 4 joule.

Explanation of the solution:

The average kinetic energy is calculated by integrating the kinetic energy function over the given time interval and dividing by the length of the interval. $KE_{avg} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} KE(t) dt$ Given $KE(t) = 3t^2$, $t_1 = 0$, $t_2 = 2$. $KE_{avg} = \frac{1}{2-0} \int_{0}^{2} 3t^2 dt = \frac{1}{2} \left[ t^3 \right]_{0}^{2} = \frac{1}{2} (2^3 - 0^3) = \frac{1}{2} (8) = 4 \text{ J}$