Problem 9 - Applied Mathematics

Question

A particle P which is moving along a straight line with a constant acceleration of \(0.3 \mathrm{~ms}^{-2}\) passes a point \(\mathrm{A}\) on the line with a velocity of \(20 \mathrm{~ms}^{-1}\). At the time when P passes A a second particle \(Q\) is \(20 \mathrm{~m}\) behind \(A\) and is moving with a constant velocity of \(30 \mathrm{~ms}^{-1}\). Prove that the particles collide.

Step-by-Step Solution

Answer

Particles P and Q collide because they have a common position at time \(t\).

Step 1: Identify Initial Conditions and Variables

Let's denote the initial position of particle P at point A as \(x_P = 0\). The velocity of particle P at this position is \(u_P = 20 \text{ ms}^{-1}\) and it has a constant acceleration \(a_P = 0.3 \text{ ms}^{-2}\). Particle Q is initially \(x_Q = -20 \text{ m}\) behind A and moves with a constant velocity \(u_Q = 30 \text{ ms}^{-1}\).

Step 2: Write Position Equations for Particles

The position of P after time \(t\) can be given by the equation \[ x_P(t) = u_P t + \frac{1}{2} a_P t^2 \]. The position of Q after time \(t\) can be given by the equation \[ x_Q(t) = x_{Q_0} + u_Q t \] where \(x_{Q_0} = -20 \text{ m}\). Therefore, \[ x_Q(t) = -20 + 30t \].

Step 3: Set Equations Equal to Each Other

For particles P and Q to collide, their positions at time \(t\) must be equal, i.e., \[ x_P(t) = x_Q(t) \]. Substitute the equations from Step 2: \[ 20t + \frac{1}{2}(0.3)t^2 = -20 + 30t \].

Step 4: Solve the Quadratic Equation

First, simplify the equation: \[ 20t + 0.15t^2 = -20 + 30t \]. Move all terms to one side to form a standard quadratic equation: \[ 0.15t^2 - 10t + 20 = 0 \]. To solve for \(t\), we use the quadratic formula \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], with \(a = 0.15\), \(b = -10\), and \(c = 20\).

Step 5: Calculate the Discriminant and Roots

The discriminant (\text{Δ}) is given by \[ b^2 - 4ac \]. First, find the values \[ b^2 = 100 \text{, and } 4ac = 4(0.15)(20) = 12\]. Therefore, \[ \text{Δ} = 100 - 12 = 88 \]. Now, calculate the roots: \[ t = \frac{10 \pm \sqrt{88}}{0.3} \]. Simplify the roots: \[ t = \frac{10 \pm 2\sqrt{22}}{0.3} \].

Key Concepts

Constant Acceleration

In physics, acceleration is the rate at which an object's velocity changes over time. When this acceleration is constant, it means the velocity of the object is changing at a uniform rate. For instance, the problem with particle P has it moving along a straight line with a constant acceleration of 0.3 m/s². This means for every second, P increases its velocity by 0.3 m/s.
One of the fundamental equations of motion for constant acceleration is:
\( v = u + at \)
where:

• \(v\) is the final velocity
• \(u\) is the initial velocity
• \(a\) is the constant acceleration
• \(t\) is the time interval

Understanding constant acceleration helps predict future motion based on an object's current state.

Kinematic Equations

Kinematics is the branch of mechanics that describes the motion of objects without considering the causes of that motion. The kinematic equations are used to relate the initial and final velocities, acceleration, time, and position of moving objects. For constant acceleration, these equations are particularly useful. The two relevant kinematic equations from the problem are:
\[ x = ut + \frac{1}{2}at^2 \]
and
\[ x = x_{0} + ut \]
where:

• The first equation is for particle P, accounting for constant acceleration
• The second equation is for particle Q, moving with a constant velocity
• \(x\) is the final position
• \(x_{0}\) is the initial position

Using kinematic equations, we can determine at what time and position two moving objects, such as particles P and Q, will collide.

Quadratic Equation

A quadratic equation is a second-order polynomial equation in a single variable. It has the general form:
\[ ax^2 + bx + c = 0 \]
where:

• \(a\), \(b\), and \(c\) are constants
• \(x\) is the variable

In our problem, we need to solve a quadratic equation to find the time \(t\) when particles P and Q collide. The derived quadratic equation is:
\[ 0.15t^2 - 10t + 20 = 0 \]
We solve this equation using the quadratic formula:
\[ t = \frac{-b \'pm \sqrt{b^2 - 4ac}}{2a} \]
where:

• \(a = 0.15\)
• \(b = -10\)
• \(c = 20\)

The discriminant \(\Delta\) helps determine the nature of the roots by calculating \(\Delta = b^2 - 4ac\). For our problem, the discriminant is positive, indicating two real and distinct solutions for time \(t\). Solving this gives us the precise moments when particles P and Q meet.