Particle Motion: Relating Acceleration and Velocity

This solution explains how to find the relationship between the acceleration of a particle and its velocity, given its position ($x$) described as a function of time ($t$) by the equation $x = (2t+3)^{-1/2}$. We will use calculus (differentiation) to find the velocity and acceleration, and then determine their proportionality.

Calculating Particle Velocity

Velocity ($v$) is the rate of change of distance ($x$) with respect to time ($t$). We find it by differentiating the position equation with respect to time:

Given: $x = (2t+3)^{-1/2}$

Using the chain rule for differentiation, let $u = 2t+3$. Then $x = u^{-1/2}$.

The derivative of $x$ with respect to $u$ is $\frac{dx}{du} = -\frac{1}{2} u^{-3/2}$.

The derivative of $u$ with respect to $t$ is $\frac{du}{dt} = 2$.

Velocity $v = \frac{dx}{dt} = \frac{dx}{du} \cdot \frac{du}{dt}$.

Substituting the derivatives:

$v = (-\frac{1}{2} u^{-3/2}) \cdot (2) = -u^{-3/2}$

Replacing $u$ with $(2t+3)$:

$v = -(2t+3)^{-3/2}$

Determining Particle Acceleration

Acceleration ($a$) is the rate of change of velocity ($v$) with respect to time ($t$). We find it by differentiating the velocity equation:

Given: $v = -(2t+3)^{-3/2}$

Using the chain rule again, let $u = 2t+3$. Then $v = -u^{-3/2}$.

The derivative of $v$ with respect to $u$ is $\frac{dv}{du} = -(-\frac{3}{2}) u^{-5/2} = \frac{3}{2} u^{-5/2}$.

The derivative of $u$ with respect to $t$ is $\frac{du}{dt} = 2$.

Acceleration $a = \frac{dv}{dt} = \frac{dv}{du} \cdot \frac{du}{dt}$.

Substituting the derivatives:

$a = (\frac{3}{2} u^{-5/2}) \cdot (2) = 3u^{-5/2}$

Replacing $u$ with $(2t+3)$:

$a = 3(2t+3)^{-5/2}$

Establishing the Acceleration-Velocity Relationship

Our goal is to find how acceleration ($a$) is proportional to velocity ($v$). We need to eliminate time ($t$) or the term $(2t+3)$ from our equations.

We have:

• $v = -(2t+3)^{-3/2}$
• $a = 3(2t+3)^{-5/2}$

From the velocity equation, we can isolate the term $(2t+3)^{-3/2}$:

$(2t+3)^{-3/2} = -v$

Now, let's express the term $(2t+3)^{-5/2}$ needed for acceleration using the velocity term. Let $Y = (2t+3)$.

We have $v = -Y^{-3/2}$ and $a = 3Y^{-5/2}$.

From $v = -Y^{-3/2}$, we raise both sides to the power of $-2/3$ to find $Y$ in terms of $v$:

$(Y^{-3/2})^{-2/3} = (-v)^{-2/3}$

$Y = (-v)^{-2/3}$

Now, substitute this expression for $Y$ into the acceleration equation $a = 3Y^{-5/2}$:

$a = 3 [(-v)^{-2/3}]^{-5/2}$

Using the exponent rule $(x^m)^n = x^{m \cdot n}$:

$a = 3 (-v)^{(-2/3) \cdot (-5/2)}$

$a = 3 (-v)^{5/3}$

Since the power $5/3$ involves an odd root, we can evaluate $(-1)^{5/3}$:

$(-v)^{5/3} = (-1)^{5/3} \cdot v^{5/3} = (-1) \cdot v^{5/3} = -v^{5/3}$

Substituting this back into the equation for $a$:

$a = 3 (-v^{5/3})$

$a = -3 v^{5/3}$

Conclusion on Proportionality

The derived relationship $a = -3 v^{5/3}$ shows that the acceleration ($a$) is directly proportional to the velocity ($v$) raised to the power of $5/3$. The negative sign indicates that the acceleration is in the opposite direction to the velocity for the given motion described by the equation.

Therefore, the acceleration is proportional to $(Velocity)^{5/3}$.