Problem 8 - Applied Mathematics

Question

A particle P starts from rest from a point \(A\) and moves along a straight line with a constant acceleration of \(2 \mathrm{~ms}^{-2}\). At the same time a second particle \(Q\) is \(5 \mathrm{~m}\) behind \(\mathrm{A}\) and is moving in the same direction as \(\mathrm{P}\) with a speed of \(5 \mathrm{~ms}^{-1}\). If \(Q\) has a constant acceleration \(3 \mathrm{~ms}^{-2}\) find how far from \(A\) it overtakes \(P\).

Step-by-Step Solution

Answer

Q overtakes P at a distance of \(60 + 10 \sqrt{35}\) meters.

Step 1 - Write the equations of motion

The equation of motion for a particle with constant acceleration is given by: \[ s = ut + \frac{1}{2}at^2 \] Where: - s is the displacement - u is the initial velocity - t is the time - a is the acceleration

Step 2 - Determine the equations for particle P

Particle P starts from rest, so the initial velocity (u) is 0. The acceleration (a) is given as \(2 \mathrm{~ms}^{-2}\). The displacement of P after time t can be written as: \[ s_P = 0 \cdot t + \frac{1}{2} \cdot 2 \cdot t^2 \] Simplified, this gives: \[ s_P = t^2 \]

Step 3 - Determine the equations for particle Q

Particle Q starts 5 meters behind A, with an initial velocity of \(5 \mathrm{~ms}^{-1}\) and an acceleration of \(3 \mathrm{~ms}^{-2}\). The displacement of Q after time t can be written as: \[ s_Q = 5 + 5t + \frac{1}{2} \cdot 3 \cdot t^2 \] Simplified, this gives: \[ s_Q = 5 + 5t + \frac{3}{2}t^2 \]

Step 4 - Set displacements equal to find the time they are at the same position

To find when Q overtakes P, set \(s_P\) equal to \(s_Q\): \[ t^2 = 5 + 5t + \frac{3}{2}t^2 \] Rearrange the equation: \[ t^2 - \frac{3}{2}t^2 - 5t - 5 = 0 \] \[ \frac{1}{2} t^2 - 5t - 5 = 0 \]

Step 5 - Solve the quadratic equation for t

To solve the quadratic equation \( \frac{1}{2} t^2 - 5t - 5 = 0 \), use the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For the equation \(\frac{1}{2}t^2 - 5t - 5 = 0\), \(a = \frac{1}{2}\), \(b = -5\), and \(c = -5\). Plugging in these values: \[ t = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(\frac{1}{2})(-5)}}{2(\frac{1}{2})} \]\[ t = \frac{5 \pm \sqrt{25 + 10}}{1} \]\[ t = 5 \pm \sqrt{35} \]We discard the negative value as time cannot be negative: \[ t = 5 + \sqrt{35} \]

Step 6 - Calculate the displacement

Using the time calculated, find the displacement when Q overtakes P: \[ s_P = (5 + \sqrt{35})^2 \] \[ s_P = 25 + 10 \sqrt{35} + 35 \] \[ s_P = 60 + 10 \sqrt{35} \]

Key Concepts

Equations of Motion

In physics, the equations of motion are used to describe the behavior of objects in motion. They establish relationships between displacement, initial velocity, acceleration, and time. The most common equation of motion for constant acceleration is:\[ s = ut + \frac{1}{2}at^2 \]

• Displacement (s): The total distance traveled by the object.
• Initial Velocity (u): The velocity at which the object starts moving.
• Acceleration (a): The rate at which the velocity of the object changes.
• Time (t): The time duration for which the object has been in motion.

These equations help predict how far an object will travel and how fast it will be moving at any given point in time when acceleration is uniform.

Initial Velocity

Initial velocity is an essential concept in motion problems. It refers to the speed and direction at which an object begins its movement. For example: In the provided exercise, particle P starts from rest, so its initial velocity (u) is 0.Here's why initial velocity matters:

• It determines the starting point of speed for the object.
• Affects the overall displacement and final speed of the object.

If the initial velocity is zero, the object starts its motion purely under the influence of acceleration. If there's an initial velocity, it's combined with the acceleration to influence the total distance covered over time.

Displacement

Displacement specifies how far an object moves from its original position to its final position. In the context of the equation of motion, it's given by:\[ s = ut + \frac{1}{2}at^2 \]

• For particle P: Since it starts from rest (u=0) and has a constant acceleration of 2 \(\mathrm{m/s^2}\), the displacement over time t is calculated as \(s_P = t^2\).
• For particle Q: It starts 5 meters behind point A with an initial velocity of 5 \(\mathrm{m/s}\) and an acceleration of 3 \(\mathrm{m/s^2}\), given by \(s_Q = 5 + 5t + \frac{3}{2}t^2\).

By comparing the displacements, we determine when particle Q overtakes particle P. At the point where \(s_P = s_Q\), the two positions coincide, giving us further understanding about their motion paths.

Quadratic Equation

Quadratic equations frequently appear in motion problems due to the squared term in the equations of motion. A quadratic equation takes the form: \[ ax^2 + bx + c = 0 \]In our example, we derive the quadratic equation by setting the displacements of particles P and Q equal:\[ t^2 = 5 + 5t + \frac{3}{2}t^2 \] When simplified, it becomes:\[ \frac{1}{2}t^2 - 5t - 5 = 0 \] To solve it, we use the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = \frac{1}{2}\), \(b = -5\), and \(c = -5\). This helps us find the precise time when the two particles will overtake each other. The solution that is a positive value of t represents the time at which this event happens, as time cannot be negative.