Question

A cartasian equation of a projectile is given by \(\mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^{2}\) Calculate its initial velocity. (A) \(\sqrt{10 \mathrm{~ms}^{-1}}\) (B) \(\sqrt{5 \mathrm{~ms}^{-1}}\) (C) \(\sqrt{2} \mathrm{~ms}^{-1}\) (D) \(4 \mathrm{~ms}^{-1}\)

Short answer

The initial velocity of the projectile is \(\sqrt{5}\,\text{ms}^{-1}\), which corresponds to option (B).

Step-by-step solution

Find the initial vertical and horizontal velocities

Since the Cartesian equation of a projectile is given by \(y = 2x - 5x^2\), we can first find the initial vertical velocity by taking the derivative of it with respect to time. We know that the horizontal velocity remains constant throughout the projectile motion, so: \[v_x = \frac{dx}{dt}\] The vertical velocity, on the other hand, changes with time and the vertical position: \[v_y = \frac{dy}{dt} = \frac{d(2x - 5x^2)}{dt}\] To find \(v_y\), we can first find the derivative of \(y\) with respect to \(x\): \[\frac{dy}{dx} = \frac{d(2x - 5x^2)}{dx} = 2 - 10x\] Then, we can find the initial vertical velocity when x = 0: \[v_{y0} = 2 - 10(0) = 2\] And since the horizontal velocity is constant, the initial horizontal velocity is: \[v_{x0} = \frac{dx}{dt} = \frac{2}{2} = 1\]

Find the initial velocity magnitude

Now that we have the initial vertical and horizontal velocities, we can use the Pythagorean theorem to find the magnitude of the initial velocity: \[v_0 = \sqrt{v_{x0}^2 + v_{y0}^2} = \sqrt{1^2 + 2^2} = \sqrt{1+4} = \sqrt{5}\] Thus, the initial velocity of the projectile is: \(\boxed{\sqrt{5}\,\text{ms}^{-1}}\), which corresponds to option (B).

Key concepts

Initial Velocity Calculation

When dealing with projectile motion, calculating the initial velocity is often the first step to understanding the object's trajectory. The initial velocity (\(v_0\)) is a vector quantity that describes the speed and direction of the projectile at the moment it is launched.

• It is composed of two parts: the initial horizontal velocity (\(v_{x0}\)) and the initial vertical velocity (\(v_{y0}\)).
• These components can be found using known equations of motion and initial conditions specific to the problem.
• The overall initial velocity magnitude can be calculated using the Pythagorean theorem: \[v_0 = \sqrt{v_{x0}^2 + v_{y0}^2}\]

For the given exercise, the projectile’s motion is described by the equation \(y = 2x - 5x^2\). To find the magnitudes of \(v_{x0}\) and \(v_{y0}\), you first differentiate this equation with respect to time. In this example, we found that \(v_{x0} = 1\) and \(v_{y0} = 2\). The calculation of the initial velocity magnitude is then \(\sqrt{5}\,\text{ms}^{-1}\).

Cartesian Equation of Motion

The Cartesian equation of motion describes how the position of a projectile changes over time in a coordinate system. In this exercise:

• We've been given the equation \(y = 2x - 5x^2\).
• This equation is a parabola, showing a curved path in a two-dimensional plane.
• The terms \(2x\) and \(-5x^2\) describe the linear and quadratic relations of the projectile's vertical displacement with horizontal displacement.

The first step in analyzing such equations is often finding the derivatives to understand the rate of change. This is essential for determining velocities or accelerations. In our specific example, \(\frac{dy}{dx} = 2 - 10x\) represents the derivative and is crucial to finding the velocity components. How each part of the equation contributes to the motion can help us make sense of the projectile dynamics.

Derivative in Physics

In physics, derivatives are vital for understanding how quantities change over time or space. Derivatives can:

• Indicate rates of change, which is invaluable for calculating velocities and accelerations.
• Help to transform position equations into velocity equations, by differentiating position with respect to time.
• Link back to fundamental physics equations and concepts.

In the context of the given exercise, finding the derivative of the Cartesian equation \(y = 2x - 5x^2\) plays a key role. By differentiating with respect to \(x\), we found \(\frac{dy}{dx} = 2 - 10x\). Substituting any known values of \(x\) can then give specific slopes, which are related to the projectile's vertical velocity. Understanding the derivatives allows students to bridge position data and kinematic equations effectively.