Problem 7 - Applied Mathematics

Question

A particle P moves along the \(x\)-axis and a particle \(Q\) moves along the \(y\)-axis. P starts from rest at the origin and moves with a constant acceleration of \(2 \mathrm{~ms}^{-2}\). At the same time \(Q\) is at the point \((0,3)\) with a velocity of \(2 \mathrm{~ms}^{-1}\) and is moving with a constant acceleration of \(-3 \mathrm{~ms}^{-2} .\) Find the distance between \(\mathrm{P}\) and \(Q\) 4 seconds later.

Step-by-Step Solution

Answer

The distance between P and Q after 4 seconds is 20.62 meters.

Step 1: Determine the position of particle P

Particle P starts from rest at the origin with a constant acceleration of 2 m/s². The position of particle P after time t can be found using the equation: \[ x_P = \frac{1}{2} a t^2 \] Here, a is the acceleration, t is the time. Plug in the given values: \[ x_P = \frac{1}{2} \times 2 \times 4^2 = 16 \text{ meters} \]

Step 2: Determine the position of particle Q

Particle Q starts at the point (0, 3) with an initial velocity of 2 m/s and a constant acceleration of -3 m/s². The position of particle Q after time t can be found using the equation: \[ y_Q = y_0 + v_0 t + \frac{1}{2} a t^2 \] Here, y_0 is the initial position, v_0 is initial velocity, a is the acceleration, and t is the time. Plug in the given values: \[ y_Q = 3 + 2 \times 4 + \frac{1}{2} \times (-3) \times 4^2 = 3 + 8 - 24 = -13 \text{ meters} \]

Step 3: Calculate the distance between P and Q

The distance between P and Q can be calculated using the distance formula: \[ d = \sqrt{(x_P - x_Q)^2 + (y_P - y_Q)^2} \] Since x_Q remains zero (Q moves along the y-axis) and x_P = 16 meters, y_P = 0 meters, y_Q = -13 meters, the distance is: \[ d = \sqrt{(16 - 0)^2 + (0 - (-13))^2} = \sqrt{16^2 + 13^2} = \sqrt{256 + 169} = \sqrt{425} = 20.62 \text{ meters} \]

Key Concepts

constant acceleration

In physics, constant acceleration means an object's velocity is changing at a steady rate over time. For our particle P, this means the object is speeding up by the same amount every second. This kind of uniform acceleration is often seen in free-fall problems, where gravity provides a constant acceleration.

position equations

To find out where a particle will be after a certain time period under constant acceleration, we can use position equations. For particle P moving from rest with acceleration and particle Q starting from a known position with both initial velocity and acceleration, we need different equations.
For particle P, starting from rest with constant acceleration:\[ x_P = \frac{1}{2} a t^2 \]
Where:

• \ta = acceleration (2 m/s² for P)
• \tt = time (in seconds)

For particle Q, starting with an initial velocity and position, the formula is:

\[ y_Q = y_0 + v_0 t + \frac{1}{2} a t^2 \]
Where:

• \ty_0 = initial position (3 meters for Q)
• \tv_0 = initial velocity (2 m/s for Q)
• \ta = acceleration (-3 m/s² for Q)
• \tt = time (in seconds)

Plugging in values helps us find the positions after a certain time.

distance formula

Once we have the positions of particles P and Q, we need to find out how far apart they are. This is where the distance formula in a coordinate system comes in.

For two points \((x_1, y_1)\) and \((x_2, y_2)\):

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

In our case, particle P has moved along the x-axis, and particle Q along the y-axis. At t = 4 seconds:

• \tx_P = 16 meters
• \ty_Q = -13 meters

Using the distance formula:\[ d = \sqrt{(16 - 0)^2 + (0 - (-13))^2 } = \sqrt{256 + 169} = \sqrt{425} \approx 20.62 \text{ meters} \]

This result gives us the separation distance between particles P and Q after 4 seconds.