Explore: Physics Applications of Rational Equations Solutions

Physics Applications of Rational Equations Solutions

In physics, you often begin with the definition of a term. Then you apply solving techniques to create a new equation by isolating a particular variable.

The average velocity, v, can be defined as the change in displacement, $∆ x$ , divided by time, t: $v = \frac{∆ x}{t}$
- v = average velocity  
- $∆ x$ = change in displacement/position
- t = time  
Assuming constant acceleration, the average velocity can also be defined as the sum of the final and initial velocities divided by two: $v = \frac{v_{f} + v_{i}}{2}$  
- $v_{f} =$ final velocity
- $v_{i} =$ initial velocity

- $v = \frac{v_{f} + v_{i}}{2}$

$\begin{array}{rcl} _{} v & = & average velocity \\ v_{f} & = & final velocity \\ v_{i} & = & initial velocity \end{array}$

- $\frac{∆ x}{t} = \frac{v_{f} + v_{i}}{2}$

Since both definitions are the average velocity, we can set these two rational equations equal to each other.

- $t = \frac{2 ∆ x}{v_{f} + v_{i}}$

Velocity equation in terms of time, t.

Acceleration can be defined as the change in velocity over time: $a = \frac{∆ v}{t}$

The change in velocity is defined as: $∆ v = v_{f} - v_{i}$

- $a = \frac{v_{f} - v_{i}}{t}$

$v_{f} =$ final velocity

$v_{i} =$ initial velocity

t = time

- $t = \frac{v_{f} - v_{i}}{a}$

Acceleration equation in terms of time, t.

Since both equations are equal to t they can be set equal to each other to obtain a new formula: $\frac{2 ∆ x}{v_{f} + v_{i}} = \frac{v_{f} - v_{i}}{a}$

Note

For these equations, assume the initial time is zero, so the change in time would be just t.

Often in physics, you would solve for the final velocity squared, $v_{f}^{2}$ . However, for the purpose of this lesson, the equation will remain in rational (fraction) form.

Example 7

Mark is driving at 20 meters per second (20 m/s) and sees an obstruction in the road. He then slams on the brakes and decelerates at 6 meters per second squared (–6 m/ $s^{2}$ ) until the car is stopped.

Calculate the distance, $∆ x$ , traveled until the car stops using the formula: $\frac{2 ∆ x}{v_{f} + v_{i}} = \frac{v_{f} - v_{i}}{a}$

Plan
Identify the given values

Solve for $∆ x$

Implement

$\begin{array}{rcl} v_{i} & = & 20 \frac{m}{s} \\ v_{f} & = & 0 \frac{m}{s (Mark comes to a full stop)} \\ a & = & - 6 \frac{m}{s^{2}} \end{array}$

$\frac{2 ∆ x}{0 + 20} = \frac{0 - 20}{- 6}$

$\begin{array}{rcl} \frac{2 ∆ x}{20} & = & \frac{- 20}{- 6} \\ \frac{∆ x}{10} & = & \frac{10}{3} \\ 3 ∆ x & = & 100 \\ ∆ x & = & 33.3 meters \end{array}$

Note

Explain 

Substitution 
Simplify (right side)
Simplify (the fractions on both sides, GCF of all terms is 2)
Cross product 
Solve

Explain
Mark will travel approximately 33.3 meters before the car stops.

Example 8

A car traveling at 15 meters per second (15 m/s) accelerates at 1 meter per second squared (1 m/ $s^{2}$ ) for a distance of 32 meters.

Find the final velocity using the formula: $\frac{2 ∆ x}{v_{f} + v_{i}} = \frac{v_{f} - v_{i}}{a}$

Plan 
Identify the given values

Solve for $v_{f}$

Implement

$∆ x = 32 meters v_{i} = 15 \frac{m}{s} a = 1 \frac{m}{s^{2}}$

$\begin{array}{rcl} \frac{2 (32)}{v_{f} + 15} & = & \frac{v_{f} - 15}{1} \\ (v_{f} + 15) (v_{f} - 15) & = & 64 \\ v_{f}^{2} - 225 & = & 64 \end{array}$

$\begin{array}{rcl} v_{f}^{2} - 289 & = & 0 \\ (v_{f} - 17) (v_{f} + 17) & = & 0 \\ v_{f} & = & \pm 17 \\ v_{f} & = & 17 \frac{m}{s} \end{array}$

Note

Explain

Substitute each value into the equation 
Cross product 
Distribute 
Set the equation equal to zero 
Factor (Difference of two squares) 
Solve 
Check for extraneous solutions

Explain

The final velocity of the car was 17 m/s.

Note

The value –17 is extraneous because in physics it denotes the opposite direction. The car did not change direction while accelerating.