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Adding and subtracting integers Multiplying and dividing integers Fractions Decimals Negative numbers Math formulas Substitution Solving equations Rearranging equations Speed distance time Quadratic equationHere you will learn about the kinematic equations, including what they are and how to use them to solve kinematics problems.

Students will first learn about kinematic equations as part of algebra in high school.

**Kinematic equations** are used to solve problems involving moving objects.

Kinematics is the math concerned with the movement of objects.

To do this, you need to know the 5 different kinematic formulas. They involve 5 different variables of motion. Velocity is the speed in a given direction. Displacement is the distance from the original position. In all kinematics formulas, the kinematic variables represent:

s= displacement

u= initial velocity

v= final velocity

a= acceleration

t= time

The kinematics formulas are,

v=u+at s=ut+\cfrac{1}{2} \, at^2 s=vt-\cfrac{1}{2} \, at^2 s=\cfrac{1}{2} \, (u+v)t v^2=u^2+2asKinematic equations can derive one or more variables when the others are known. These equations describe motion under constant velocity or constant acceleration.

Since they only apply to situations with unchanging acceleration or speed, they cannot be used when either is varying.

The kinematic equations or kinematic formulas are sometimes known as the suvat equations or equations of motion.

These kinematic formulas are used when there is constant acceleration. You do not need to worry about other factors affecting the motion of an object such as air resistance.

The kinematic formulas are based on Newtonβs laws of motion. You may also come across these formulas in high school physics problems.

When an object is slowing down, it has a negative acceleration. This is known as deceleration.

How does this relate to high school math?

**High School – Algebra – Creating Equations (HS.A.CED.A.4)**Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. For example, rearrange Ohm’s law V=IR to highlight resistance R.

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DOWNLOAD FREEIn order to work with kinematic equations:

**Use the formula given in the question.****Work carefully to answer the question, one step at a time.****Write the final answer clearly.**

Given that s=\cfrac{1}{2} \, (u+v)t, find s, when u=5, \, v=20 and t=4.

**Use the formula given in the question.**

The formula is s=\cfrac{1}{2} \, (u+v)t .

2**Work carefully to answer the question, one step at a time.**

Substitute the known values into the formula.

\begin{aligned}&s=\cfrac{1}{2} \, (u+v)t \\\\ &s=\cfrac{1}{2}\times(5+20)\times{4} \\\\ &s=50 \end{aligned}3**Write the final answer clearly.**

The final answer is s=50.

Rearrange v=u+at to make a the subject.

**Use the formula given in the question.**

The formula is v=u+at.

**Work carefully to answer the question, one step at a time.**

You need to rearrange the formula, one step at a time using inverse operations.

\begin{aligned}&v=u+at \\\\ &v-u=at \\\\ &\cfrac{v-u}{t}=a \end{aligned}

**Write the final answer clearly.**

The final answer is a=\cfrac{v-u}{t}.

For the following question, use the formula v^2=u^2+2as.

Calculate s when v is 10, \, u is 0 and a is 20.

**Use the formula given in the question.**

The formula given in the question is v^2=u^2+2as.

**Work carefully to answer the question, one step at a time.**

First, substitute the values into the kinematic formula:

\begin{aligned}v^2&=u^2+2as \\\\
10^2&=0^2+2\times{20}\times{s} \\\\
100&=0+40s \end{aligned}

Then you can rearrange to solve for s.

\begin{aligned}100&=0+40s \\\\ 100&=40s \\\\ \cfrac{100}{40}&=s \\\\ 2.5&=s \end{aligned}

**Write the final answer clearly.**

The final answer is s=2.5.

For the following question, you may use the formula s=ut+\cfrac{1}{2} \, at^{2} , where

t= time \hspace{1.5cm} u= initial velocity

a= acceleration \hspace{0.5cm} s= displacement

A car is stopped at a traffic light. When the light goes green, the car immediately accelerates at 5\mathrm{~m/s}^{2}. How far does the car travel in 10 seconds?

**Use the formula given in the question.**

The formula given is s=ut+\cfrac{1}{2} \, at^2.

**Work carefully to answer the question, one step at a time.**

The initial velocity of the object u is 0 as the car is stationary at the start of the kinematics situation. You substitute the values into the kinematic formula.

\begin{aligned}s&=ut+\cfrac{1}{2} \, at^{2} \\\\
s&=0\times{10}+\cfrac{1}{2}\times{5}\times{10}^{2} \\\\
s&=250 \end{aligned}

**Write the final answer clearly.**

The final answer is, the car has traveled 250\mathrm{~m} in 10 seconds.

For the following question, you may use the formula s=vt-\cfrac{1}{2} \, at^{2} , where

t= time \hspace{1.5cm} v= final velocity

a= acceleration \hspace{0.5cm} s= displacement

A ball was dropped from stationary and accelerated at 10\mathrm{~m/s}^{2}. The ball passed a motion sensor after 15\mathrm{~s}, travelling at 90\mathrm{~m/s}. How far was the motion sensor away from the initial position of the ball?

**Use the formula given in the question.**

The formula given is s=vt-\cfrac{1}{2} \, at^2.

**Work carefully to answer the question, one step at a time.**

The final velocity of the object v=90\mathrm{~m/s} with t=15\mathrm{~s} and a=10\mathrm{~m/s}^{2}. You substitute the values into the kinematic formula.

\begin{aligned}s&=vt-\cfrac{1}{2} \, at^{2} \\\\ s&=90\times{15}-\cfrac{1}{2}\times{10}\times{15}^{2} \\\\ s&=225 \end{aligned}

**Write the final answer clearly.**

The motion sensor is 225\mathrm{~m} below the initial position of the ball.

- Use position-time and velocity-time graphs to illustrate the relationships between the variables. This helps students see how changes in one variable affect another. Graphs make the abstract set of equations more concrete. Look for online interactive graphs to make this more engaging.

- Guide students through the derivation of kinematic equations by starting with the basic definitions of velocity and acceleration. Start with v=u+at, and then progress to the equations for displacement and average velocity.

- Once students are comfortable with basic kinematic problem solving, introduce free-fall or projectile motion problems. Show how the same equations can be applied, just with specific values for acceleration due to the gravitation on Earth.

- Focus on one-dimensional motion, where objects move along a straight path, either horizontally or vertically, without lateral movement. This simplicity helps students grasp more complex motion later, such as two-dimensional motion or rotational motion around an axis.

- For a more advanced tip, introduce angular acceleration after students understand linear acceleration. Explain how angular acceleration relates to rotational motion, where the rate of change of angular velocity mirrors linear acceleration in circular paths.

**Not knowing acceleration can be negative**

You may be asked to work out the acceleration of a situation involving the motion of an object. Acceleration can be negative. When this happens the object is slowing down. This is known as deceleration.

**Mixing up \textbf{v} and \textbf{u}**

The variables v and u can easily be muddled as they look similar and they both represent velocity. u is the initial velocity; the velocity of the object at the beginning of the situation. v is the final velocity; the velocity at the end of the situation.

**Inconsistent units**

The units that are mostly used in kinematics are m (meters) for displacement, s (seconds) for time, m/s (meters per second) for velocity and m/s^{2} (meters per second squared) for acceleration. Make sure the units in your answer are consistent with those of the given values.

**Not knowing initial velocity of an object at rest**

Sometimes you are told the objectβs initial position is at rest, this means that the initial velocity u is 0\mathrm{~m/s}.

1. Given that v=u+at, find v when u=50, \, a=- \, 10, \, t=4.

10

90

30

70

Substitute the values into the equation and solve.

\begin{aligned}v&=u+at \\\\ v&=50+(- \, 10)\times{4} \\\\ v&=50-40 \\\\ v&=10 \end{aligned}

2. Given that s=\cfrac{1}{2} \, (u+v)t, calculate s when u=10, \, v=80 and t=6.

540

15

30

270

Substitute the values into the equation and solve.

\begin{aligned}s&=\cfrac{1}{2} \, (u+v)t \\\\ s&=\cfrac{1}{2}\times(10+80)\times{6} \\\\ s&=\cfrac{1}{2}\times{90}\times{6} \\\\ s&=270 \end{aligned}

3. Rearrange v^{2}=u^{2}+2as to make s the subject.

s=\cfrac{v^{2}+2a}{u^2}

s=\cfrac{v^{2}-u^{2}}{2a}

s=\cfrac{v^{2}+u^{2}}{2a}

s=\cfrac{v^{2}-2a}{u^{2}}

Rearrange, taking one step at a time.

\begin{aligned}&v^{2}=u^{2}+2as \\\\ &v^{2}-u^{2}=2as \\\\ &\cfrac{v^{2}-u^{2}}{2a}=s \end{aligned}

4. Given that v=u+at, find u when v=200, \, a=20, \, t=3.

180

300

140

360

Substitute the known values into the formula, then solve to find the unknown variable u.

\begin{aligned}v&=u+at \\\\ 200&=u+20\times{3} \\\\ 200&=u+60 \\\\ u&=140 \end{aligned}

5. A ball is dropped from a tower and is in free fall (vertical motion). Its initial velocity is 0\mathrm{~m/s}. The gravitational acceleration is 10\mathrm{~m/s}^{2}. The final velocity as the ball hits the ground is 50\mathrm{~m/s}. Calculate the time the ball was in free fall. You may use the formula v=u+at.

40 seconds

5 seconds

40 minutes

10 minutes

Substitute the known values into the formula, then rearrange to solve to find the unknown variable t.

\begin{aligned}v&=u+at \\\\ 50&=0+10\times{t} \\\\ 50&=10t \\\\ t&=5 \end{aligned}

The answer will be </span><span style="font-weight: 400;">5</span><span style="font-weight: 400;"> seconds, as seconds is consistent with the other units.

6. A train is traveling at 30\mathrm{~m/s}. It brakes for 3 seconds and it decelerates at 5\mathrm{~m/s}^{2}. Calculate how far the train travels in those 3 seconds. You may use the formula s=ut+\cfrac{1}{2} \, at^2.

67.5\mathrm{~m}

112.5\mathrm{~m}

82.5\mathrm{~m}

97.5\mathrm{~m}

Substitute the known values into the formula.

\begin{aligned}s&=ut+\cfrac{1}{2} \, at^2 \\\\ s&=30\times 3+\cfrac{1}{2}\times(- \, 5)\times{3^2} \\\\ s&=90-22.5 \\\\ s&=67.5 \end{aligned}

The answer will be 67.5 meters, as meters is consistent with the other units.

Kinematic equations describe the motion of objects under constant acceleration. They relate variables like displacement, velocity, time intervals, and acceleration.

Displacement is the straight-line distance between the starting and ending point of an object’s motion, considering direction. Distance is the total path length traveled by the object, regardless of direction. Kinematic equations deal with displacement, not distance.

Each equation is derived from the others using basic algebra and the definition of acceleration. For example, you can derive the second equation from the first kinematic equation by substituting velocity into the formula for displacement.

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