# Slope Intercept Form – Equation of a Line Example

Slope intercept form is a function of a straight line  with the form

y = mx + b
where
m is the slope of the line and
b is the y-intercept.

Slope refers to how steep the line changes as the values of x change where the y-intercept is where the line crosses the y-axis.

Here are a pair of worked examples to show how to find the equation of a line from different initial conditions.

### Slope Intercept Form Example 1

This example shows how to find the equation of the line when given the slope and one point on the line.

Q: What is the Slope intercept form of the line with slope 23 and passes through the point (-3, -4)

Here’s the graph of that line. A slope of 23 means the line will rise 2 points for every 3 points of x to the right. If we start at the point (-3, -4) we can count 2 up and three over to see the way the line will progress.

The slope intercept form is

y = mx + b

To solve this, we need to know both the slope and the y-intercept. We know the slope, so we need to find the value of b. We know from the given:
m = 23
x = -3
y = -4

Plug these into the equation

-4 = 23 (-3) + b
-4 = -2 + b
-2 = b

Now we have all we need for the equation of the line.

y = 23 x + (-2)
y = 23 x – 2

We can see from the graph the line crosses the y-axis at -2 but it nice to have a second opinion.

### Slope Intercept Form Example 1 – Alternate Method

There is an alternate method to solve this problem where you can memorize an equation to eliminate the first couple steps. That formula is

(y – y1) = m(x – x1)

where x1 and yare the coordinates of the given point.

Let’s plug in the point (-3, -4) from above.

(y – y1) = m(x – x1)
(y – (-4)) = 23(x – (-3))
y + 4 = 23(x + 3)
y + 4 = 23x + 23(3)
y + 4 = 23x + 2
y = 23x + 2 – 4
y = 23x – 2

### Slope Intercept Form Example 2

The second example, two points of the line are given in the initial conditions.
Let’s find the same line as before. The red line passes through points (-6, -6) and (9, 4). Find the equation of the line.

First we need to find the slope.

The formula to find slope between two points (x1, y1) and (x2, y2) is

For our two points:
x1 = -6
y1 = -6
x2 = 9
y2 = 4

Plug these into the formula

m = 1015
reduce the fraction by factoring out a 5
m = 23

Now we can find the y-intercept in the same way we did above. Choose one of the points for x and y. Let’s try the (-6, -6) point.

y = mx + b
-6 = 23(-6) + b
-6 = –123 + b
-6 = -4 + b
-2 = b

Just to show it doesn’t matter which point you choose, let’s use the (9, 4) point.

y = mx + b
4 = 2(9) + b
4 = 183 + b
4 = 6 + b
-2 = b

Plug into the slope intercept formula

y = 23 x + (-2)
y = 23 x – 2

Which matches what we expected to see from the first example.

The key to this type of problem is to find the slope of the line. Once you have the slope, finding the y-intercept is easy. For more examples of finding the slope, check out the What is Slope? page and examples

# SOHCAHTOA Example Problem – Trigonometry Help

SOHCAHTOA is the mnemonic used to remember which sides of a right triangle are used to find the ratios needed to determine the sine, cosine or tangent of an angle. Here are a pair of SOHCAHTOA example problems to help show how to use these relationships. If you have no idea what SOHCAHTOA means, check out this introduction to SOHCAHTOA.

### SOHCAHTOA Example Problem 1

A mathematically inclined squirrel sits atop a 14-foot tall tree. He spies a nut on the ground some distance away. After careful measurements, he determines the nut is 74º from the base of his tree.
How far away is the nut from:
A) The base of the tree?
B) The math squirrel?

Here is a layout of the problem.

Part A: How far is the nut from the base of the tree?

Looking at our triangle, we see we know the angle and the length of the adjacent side. We want to know the length of the opposite side of the triangle. The part of SOHCAHTOA with these three parts is TOA, or tan θ = opposite / adjacent.

tan θ = opposite / adjacent
tan ( 74º ) = opposite / 14 ft

solve for ‘opposite’

opposite = 14 ft ⋅ tan ( 74º )
opposite = 14 ft ⋅ 3.487
opposite = 48.824 ft

The nut is 48.824 ft from the base of the tree.

Part B: How far is the nut from the math squirrel?

This time, the needed distance is the hypotenuse of the triangle. The part of SOHCAHTOA with  both adjacent and hypotenuse is CAH, or cos θ = adjacent / hypotenuse.

cos θ = adjacent / hypotenuse
cos ( 74º ) = 14 ft / hypotenuse
hypotenuse = 14 ft / cos ( 74º )
hypotenuse = 14 ft / 0.276
hypotenuse = 50.791 ft

We can use the Pythagorean Theorem to check our work.

a2 + b2 = c2
(14 ft)2 + (48.824 ft)2 = c2
196 ft2+ 2383.783 ft2= c2
2579.783 ft2 = c2
50.792 ft = c

The two numbers are close enough to account for rounding errors.

### SOHCAHTOA Example Problem 2

A prince has arrived to rescue a princess standing in a tower balcony 15 meters up across a moat 12 meters wide. He knew the princess was high up and there was a moat, so he brought the longest ladder in the kingdom, a 20-meter monster.

A) Is the ladder long enough?
B) If the top of the 20 m ladder touches the balcony, how far away from the tower wall is the bottom of the ladder?
C) What is the angle between the ground and the ladder?

Here is an illustration of our situation.

Part A: Is the ladder long enough?  To know if the ladder is long enough, we need to know how long it needs to be. At the very least, the prince’s ladder must reach the 15-meter balcony from the edge of the 12-meter moat. Use the Pythagorean theorem to find out how far up the tower wall a 20-meter ladder will reach.

a2 + b2 = c2
height2 = (20 m)2 – (12 m)2
height2 = 400 m2 – 144 m2
height2 = 256 m2
height = 16 m

A 20-meter ladder can reach 16 meters up the side of the tower. This is a meter longer than the 15 meters needed to reach the princess balcony.

Part B: When the ladder touches the edge of the balcony, how far away from the tower wall is the bottom of the ladder?

The first part measured the height the 20-meter ladder reached when the ladder was placed at the edge of the moat. We found we had more ladder than needed. If the ladder touches the balcony, we know the height it reaches is 15 meters. The ladder is still 20 meters long. We just need to find out how far from the tower to stick the bottom of the ladder. Use the Pythagorean theorem again.

a2 + b2 = c2
(balcony height)2 + (ground distance)2 = ladder2
(ground distance)2 = ladder2 – (balcony height)2

(ground distance)2 = (20 m)2 – (15 m)2
(ground distance)2 = 400 m2 – 225 m2
(ground distance)2 = 175 m2
ground distance = 13.23 m

The ladder’s base is 13.23 m away from the tower.

Part C: What is the angle between the ground and the ladder?

We were given the height of the balcony, which in this case is the ‘opposite’ side of the triangle from the needed angle. We also know the length of the ladder that forms the hypotenuse of the triangle. The part of SOHCAHTOA that has both of these parts is SOH, or sin θ = opposite/hypotenuse. Use this to solve for the angle.

sin θ = opposite/hypotenuse
sin θ = 15 m / 20 m
sin θ = 0.75
θ = sin-1(0.75)
θ = 48.59°

You can check your work in Part B if you use the answer we got as the ‘adjacent’ side of the triangle.

If you use the ladder as the hypotenuse, use CAH, or cos θ = adjacent / hypotenuse.

cos θ = adjacent / hypotenuse
cos θ = 13.23 m / 20 m
cos θ = 0.662
θ = cos-1(0.662)
θ = 48.59°

The angle matches the value above.

If you choose the balcony height as the opposite side, use TOA, or tan θ = opposite / adjacent

tan θ = opposite / adjacent
tan θ = 15 m / 13.23 m
tan θ = 1.134
θ = tan-1(1.134)
θ = 48.59°

Pretty neat how it doesn’t really matter which sides of the right triangle you use, as long as you use the correct part of SOHCAHTOA.

# Vertical Motion Example Problem – Coin Toss Equations of Motion

This equations of motion under constant acceleration example problem shows how to determine the maximum height, velocity and time of flight for a coin flipped into a well. This problem could be modified to solve any object tossed vertically or dropped off a tall building or any height. This type of problem is a common equations of motion homework problem.

Problem:
A girl flips a coin into a 50 m deep wishing well. If she flips the coin upwards with an initial velocity of 5 m/s:
a) How high does the coin rise?
b) How long does it take to get to this point?
c) How long does it take for the coin to reach the bottom of the well?
d) What is the velocity when the coin hits the bottom of the well?

Solution:
I have chosen the coordinate system to begin at the launch point. The maximum height will be at point +y and the bottom of the well is at -50 m. The initial velocity at launch is +5 m/s and the acceleration due to gravity is equal to -9.8 m/s2.

The equations we need for this problem are:

1) y = y0 + v0t + ½at2

2) v = v0 + at

3) v2 = v02 + 2a(y – y0)

Part a) How high does the coin rise?

At the top of the coin’s flight, the velocity will equal zero. With this information, we have enough to use equation 3 from above to find the position at the top.

v2 = v02 – 2a(y – y0)
0 = (5 m/s)2 + 2(-9.8 m/s2)(y – 0)
0 = 25 m2/s2 – (19.6 m/s2)y
(19.6 m/s2)y = 25 m2/s2
y = 1.28 m

Part b) How long does it take to reach the top?

Equation 2 is the useful equation for this part.

v = v0 + at
0 = 5 m/s + (-9.8 m/s2)t
(9.8 m/s2)t = 5 m/s
t = 0.51 s

Part c) How long does it take to reach the bottom of the well?

Equation 1 is the one to use for this part. Set y = -50 m.

y = y0 + v0t + ½at2
-50 m = 0 + (5 m/s)t + ½(-9.8 m/s2)t2
0 = (-4.9 m/s2)t2 + (5 m/s)t + 50 m

This equation has two solutions. Use the quadratic equation to find them.

where
a = -4.9
b = 5
c = 50

t = 3.7 s or t = -2.7 s

The negative time implies a solution before the coin was tossed. The time that fits the situation is the positive value. The time to the bottom of the well was 3.7 seconds after being tossed.

Part d) What was the velocity of the coin at the bottom of the well?

Equation 2 will help here since we know the time it took to get there.

v = v0 + at
v = 5 m/s + (-9.8 m/s2)(3.7 s)
v = 5 m/s – 36.3 m/s
v = -31.3 m/s

The velocity of the coin at the bottom of the well was 31.3 m/s. The negative sign means the direction was downward.

If you need more worked examples like this one, check out these other constant acceleration example problems.
Equations of Motion – Constant Acceleration Example Problem
Equations of Motion – Interception Example Problem
Projectile Motion Example Problem

# Friction Example Problem – Physics Homework Help

Friction is the resistance force generated between two bodies as they move across each other. It is proportional to the force that presses the two bodies together. This diagram shows the forces acting on a block sitting on a surface.

The block is pulled down onto the surface by the force of gravity while the surface pushes back in an equal and opposite force known as the normal force: N. Note there are no horizontal forces. If a horizontal force is applied such as pushing the block to the right, the block will begin to accelerate. Experience tells us this does not always happen. If you try to push something heavy, it doesn’t always move until you’ve pushed HARD enough. There must be a force working in the opposite direction of the push to resist the motion. This force is the force of friction, Ff.

Experiments have shown the magnitude of this force is dependent on the normal force. The magnitude of the friction force is directly proportional to the magnitude of the normal force. The proportionality constant between them is called the coefficient of friction, μf. The f subscript is commonly left off, and is not unusual to see just the μ listed.

The coefficient of friction depends on two factors.
The first depends on the materials the two objects are made of. It is generally easier to move 50 kg of ice on a glass surface than a 50 kg stone on sand. Each two materials have their own coefficients of friction.
The second factor is whether or not the block is moving. You may have noticed it is usually easier to move a heavy object once it is already moving. This means there are two different coefficients of friction. One for when the block is stationary, μs (static) and one for when the block begins to move, μk (kinetic).

The static coefficient is used whenever the block is stationary. As the force pushing the block increases, it will eventually reach a point where the block is on the verge of moving. The coefficient of static friction, μs is experimentally determined by carefully measuring the force at this point. The frictional force required to reach this point is F = -μsN. The minus sign indicates the direction of the force. Frictional forces oppose the force trying to move the object and act in the opposite direction. Any magnitude of force less than musN, the block will not move.

When the block is moving, the coefficient of kinetic friction is used. This value is experimentally calculated by measuring the force necessary to keep the block moving at a constant velocity. This force will equal -μkN.

Now, let’s try a friction example problem.

Example Problem:

A block weighing 200 N is pushed along a surface. If it takes 80 N to get the block moving and 40 N to keep the block moving at a constant velocity, what are the coefficients of friction μs and μk?

Solution:

For the coefficient of static friction, we need the force needed to get the block moving. In this case, 80 N.

From the description above:

Ff = μsN

N is equal to the weight of the block, so N = 200 N. Put these values into the formula.

80 N = μs·200 N
or
μs = 0.4

For the coefficient of kinetic friction, the force needed to maintain a constant velocity was 40 N. Use the formula:

Ff = μkN
40 N = μk·200 N
μk = 0.2

The two coefficients of friction for this system are μs = 0.4 and μk = 0.2.

There are two important things to remember in friction homework problems. The first is the normal force N is always perpendicular to the surface. The normal force is not always ‘up’. The second is the friction force works opposite in direction to the motion of the block. Friction is a resistive force.

# Calculate Percent Error

Percent error is the percent difference between a measured and expected value. (image: Sherman Geronimo-Tan)

Percent Error Definition

Percent error, sometimes referred to as percentage error, is an expression of the difference between a measured value and the known or accepted value. It is often used in science to report the difference between experimental values and expected values.

The formula for calculating percent error is:

Note: occasionally, it is useful to know if the error is positive or negative. If you need to know the positive or negative error, this is done by dropping the absolute value brackets in the formula. In most cases, absolute error is fine. For example, in experiments involving yields in chemical reactions, it is unlikely you will obtain more product than theoretically possible.

### Steps to Calculate the Percent Error

1. Subtract the accepted value from the experimental value.
2. Take the absolute value of step 1
3. Divide that answer by the accepted value.
4. Multiply that answer by 100 and add the % symbol to express the answer as a percentage.

Now let’s try an example problem.

You are given a cube of pure copper. You measure the sides of the cube to find the volume and weigh it to find its mass. When you calculate the density using your measurements, you get 8.78 grams/cm3. Copper’s accepted density is 8.96 g/cm3. What is your percent error?

Solution:
experimental value = 8.78 g/cm3
accepted value = 8.96 g/cm3

Step 1: Subtract the accepted value from the experimental value.

8.96 g/cm3 – 8.78 g/cm3 = -0.18 g/cm3

Step 2: Take the absolute value of step 1

|-0.18 g/cm3| = 0.18 g/cm3

Step 3: Divide that answer by the accepted value.

Step 4: Multiply that answer by 100 and add the % symbol to express the answer as a percentage.

0.02 x 100 = 2
2%

The percent error of your density calculation was 2%.

# Guidelines for Balancing Chemical Equations

One of the basic skills you will develop as you study chemistry is the ability to balance a chemical equation. One you have written a balanced equation, you will be in a good position to perform all manner of calculations! Balancing a chemical equation refers to establishing the mathematical relationship between the quantity of reactants and products. The quantities are expressed as grams or moles.

It takes practice to be able to write balanced equation. There are essentially three steps to the process:

• Write the unbalanced equation.
• Chemical formulas of reactants are listed on the lefthand side of the equation.
• Products are listed on the righthand side of the equation.
• Reactants and products are separated by putting an arrow between them to show the direction of the reaction. Reactions at equilibrium will have arrows facing both directions.
• Balance the equation.
• Apply the Law of Conservation of Mass to get the same number of atoms of every element on each side of the equation. Tip: Start by balancing an element that appears in only one reactant and product.
• Once one element is balanced, proceed to balance another, and another, until all elements are balanced.
• Balance chemical formulas by placing coefficients in front of them. Do not add subscripts, because this will change the formulas.
• Indicate the states of matter of the reactants and products.
• Use (g) for gaseous substances.
• Use (s) for solids.
• Use (l) for liquids.
• Use (aq) for species in solution in water.
• Write the state of matter immediately following the formula of the substance it describes.

### Worked Example Problem

Tin oxide is heated with hydrogen gas to form tin metal and water vapor. Write the balanced equation that describes this reaction.

1. Write the unbalanced equation.SnO2 + H2 → Sn + H2O
2. Balance the equation.Look at the equation and see which elements are not balanced. In this case, there are two oxygen atoms on the lefthand side of the equation and only one on the righthand side. Correct this by putting a coefficient of 2 in front of water:SnO2 + H2 → Sn + 2 H2O

This puts the hydrogen atoms out of balance. Now there are two hydrogen atoms on the left and four hydrogen atoms on the right. To get four hydrogen atoms on the right, add a coefficient of 2 for the hydrogen gas. Remember, coefficients are multipliers, so if we write 2 H2O it denotes 2×2=4 hydrogen atoms and 2×1=2 oxygen atoms.

SnO2 + 2 H2 → Sn + 2 H2O

The equation is now balanced. Be sure to double-check your math! Each side of the equation has 1 atom of Sn, 2 atoms of O, and 4 atoms of H.

3. Indicate the physical states of the reactants and products.To do this, you need to be familiar with the properties of various compounds or you need to be told what the phases are for the chemicals in the reaction. Oxides are solids, hydrogen forms a diatomic gas, tin is a solid, and the term ‘water vapor’ indicates that water is in the gas phase:SnO2(s) + 2 H2(g) → Sn(s) + 2 H2O(g)

There you go! This is the balanced equation for the reaction. Remember, no elements appear on one side of the reaction and not on the other. It’s always a good idea to check by counting up the number of atoms of each element just to be sure you have balanced your equation correctly. This was a straightforward example, using conservation of mass. For reactions involving ions, conservation of charge will also come into play.