Question 3

The relation between time $$t$$ and distance $$x$$ is $$t = ax^2 + bx$$ where $$a$$ and $$b$$ are constants. The acceleration is

Solution

Solution & Explanation

1. Relate Velocity to the Given Time Equation

We are given the relation between time ($$t$$) and distance ($$x$$) as:

$$t = a \cdot x^2 + b \cdot x$$

Velocity ($$v$$) is defined as the instantaneous rate of change of position with respect to time ($$v = \frac{dx}{dt}$$). To find an expression involving velocity, we differentiate both sides of the equation with respect to distance ($$x$$):

$$\frac{dt}{dx} = \frac{d}{dx}(a \cdot x^2 + b \cdot x)$$

$$\frac{dt}{dx} = 2a \cdot x + b$$

Since velocity is the reciprocal of $$\frac{dt}{dx}$$, we can write:

$$v = \frac{dx}{dt} = \frac{1}{2a \cdot x + b} \quad \text{--- (Eq. 1)}$$

2. Differentiate Velocity to Find Acceleration ($$A$$)

Acceleration is defined as the rate of change of velocity with respect to time ($$A = \frac{dv}{dt}$$). Using the chain rule, we can rewrite acceleration in terms of position ($$x$$):

$$A = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \cdot \frac{dv}{dx}$$

Let us differentiate Equation 1 with respect to $$x$$ using the power rule/reciprocal rule ($$\frac{d}{dx}(\frac{1}{u}) = -\frac{1}{u^2} \cdot \frac{du}{dx}$$):

$$\frac{dv}{dx} = \frac{d}{dx}\left[ (2a \cdot x + b)^{-1} \right]$$

$$\frac{dv}{dx} = -1 \cdot (2a \cdot x + b)^{-2} \cdot \frac{d}{dx}(2a \cdot x + b)$$

$$\frac{dv}{dx} = -\frac{2a}{(2a \cdot x + b)^2}$$

3. Substitute Velocity Back into the Acceleration Expression

From Equation 1, we know that $$\frac{1}{2a \cdot x + b} = v$$. Squaring both sides gives:

$$\frac{1}{(2a \cdot x + b)^2} = v^2$$

Substitute this back into our derivative expression for $$\frac{dv}{dx}$$:

$$\frac{dv}{dx} = -2a \cdot v^2$$

Now, calculate the final acceleration ($$A$$):

$$A = v \cdot \frac{dv}{dx} = v \cdot (-2a \cdot v^2) = -2a \cdot v^3$$

Concept Check: The negative sign confirms that as the object moves further along ($$x$$ increases), its velocity decreases over time under this specific configuration, resulting in a retarding acceleration proportional to the cube of its instantaneous speed.