The displacement of a body is given to be proportional to the cube of time elapsed. The magnitude of the acceleration of the body, is
- constant but not zero
- increasing with time
- zero
- decreasing with time
The Correct Option is B
Solution and Explanation
Given that displacement of the body,
\(y \,\propto\,t^{3}\) or\(\,\,\,\,\,y = kt^{3}\)
\(\therefore\,\,\,\,\,v = \frac{dy}{dt } = 3kt^{2}\) or \(\,\,\,\,a = \frac{dv}{dt} = 6\,kt \propto\) t
\(\therefore\,\,\,\,\,\) The magnitude of the acceleration of the body increases with time.
Therefore, the correct option is (B): increasing with time
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Concepts Used:
Rate of Change of Quantities
The rate of change of quantities can be expressed in the form of derivatives. Rate of change of one quantity with respect to another is one of the major applications of derivatives. The rate of change of a function with respect to another quantity can also be done using chain rule.
If some other quantity ‘y’ causes some change in a quantity of certain ‘x’, in view of the fact that an equation of the form y = f(x) gets always satisfied, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
This is also called the Average Rate of Change.
If the rate of change of a function is to be defined at a specific point i.e. a specific value of ‘x’, it is known as the Instantaneous Rate of Change of the function at that point. From the definition of the derivative of a function at a point, we have
From this, it is to be concluded that the instantaneous Rate of Change of the function is represented by the derivative of a function. From the rate of change formula, it represents the case when Δx → 0. Thus, the rate of change of ‘y’ with respect to ‘x’ at x = x0 = (dy/dx)x = x0