Grade 10 - Mechanics - LO.2 - kinematic equations

      

We have in LO.2 Mechanics G10

First: the Concepts

A. Area under a curve

B. Kinematic equations for 1-D motion with constant acceleration

C. free-fall motion

Second: the References

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Third: the Videos links

Fourth: Skills

A. Explain why displacement equals area under a velocity-time graph.

B. Use a velocity/time graph to explain, justify and/or derive kinematics equations for constant acceleration problems

C. Solve 1-dimensional motion problems using the kinematic equations for constant acceleration.

D. Solve kinematics problems involving vertical (free-fall) motion under gravity.

Fifth: the materials as PPT., DOCX., and PDF

In the Drive from this link

Few Notes:

What are the kinematic formulas?

The kinematic formulas are a set of formulas that relate the five kinematic variables listed below.

$\Delta x\quad\text{Displacement}$delta, x, start text, D, i, s, p, l, a, c, e, m, e, n, t, end text
$t\qquad\text{Time interval}~$t, start text, T, i, m, e, space, i, n, t, e, r, v, a, l, end text, space
$v_0 ~~\quad\text{Initial velocity}~$v, start subscript, 0, end subscript, space, space, start text, I, n, i, t, i, a, l, space, v, e, l, o, c, i, t, y, end text, space
$v\quad ~~~\text{Final velocity}~$v, space, space, space, start text, F, i, n, a, l, space, v, e, l, o, c, i, t, y, end text, space
$a \quad~~ \text{ Constant acceleration}~$a, space, space, start text, space, C, o, n, s, t, a, n, t, space, a, c, c, e, l, e, r, a, t, i, o, n, end text, space

If we know three of these five kinematic variables—$\Delta x, t, v_0, v, a$delta, x, comma, t, comma, v, start subscript, 0, end subscript, comma, v, comma, a—for an object under constant acceleration, we can use a kinematic formula, see below, to solve for one of the unknown variables.

The kinematic formulas are often written as the following four equations. 

$\Large 1. \quad v=v_0+at$1, point, v, equals, v, start subscript, 0, end subscript, plus, a, t

$\Large 2. \quad {\Delta x}=(\dfrac{v+v_0}{2})t$2, point, delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t

$\Large 3. \quad \Delta x=v_0 t+\dfrac{1}{2}at^2$3, point, delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared

$\Large 4. \quad v^2=v_0^2+2a\Delta x$4, point, v, squared, equals, v, start subscript, 0, end subscript, squared, plus, 2, a, delta, x

Since the kinematic formulas are only accurate if the acceleration is constant during the time interval considered, we have to be careful to not use them when the acceleration is changing. Also, the kinematic formulas assume all variables are referring to the same direction: horizontal $x$x, vertical $y$y, etc. 

What is a freely flying object—i.e., a projectile?

It might seem like the fact that the kinematic formulas only work for time intervals of constant acceleration would severely limit the applicability of these formulas. However one of the most common forms of motion, free fall, just happens to be constant acceleration.

All freely flying objects—also called projectiles—on Earth, regardless of their mass, have a constant downward acceleration due to gravity of magnitude $g=9.81\dfrac{\text{m}}{\text{s}^2}$g, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction.

$\Large g=9.81\dfrac{\text{m}}{\text{s}^2}\quad \text{(Magnitude of acceleration due to gravity)}$g, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, start text, left parenthesis, M, a, g, n, i, t, u, d, e, space, o, f, space, a, c, c, e, l, e, r, a, t, i, o, n, space, d, u, e, space, t, o, space, g, r, a, v, i, t, y, right parenthesis, end text

A freely flying object is defined as any object that is accelerating only due to the influence of gravity. We typically assume the effect of air resistance is small enough to ignore, which means any object that is dropped, thrown, or otherwise flying freely through the air is typically assumed to be a freely flying projectile with a constant downward acceleration of magnitude $g=9.81\dfrac{\text{m}}{\text{s}^2}$g, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction.

This is both strange and lucky if we think about it. It's strange since this means that a large boulder will accelerate downwards with the same acceleration as a small pebble, and if dropped from the same height, they would strike the ground at the same time. 

It's lucky since we don't need to know the mass of the projectile when solving kinematic formulas since the freely flying object will have the same magnitude of acceleration, $g=9.81\dfrac{\text{m}}{\text{s}^2}$g, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, no matter what mass it has—as long as air resistance is negligible.

Note that $g=9.81\dfrac{\text{m}}{\text{s}^2}$g, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction is just the magnitude of the acceleration due to gravity. If upward is selected as positive, we must make the acceleration due to gravity negative $a_y=-9.81\dfrac{\text{m}}{\text{s}^2}$a, start subscript, y, end subscript, equals, minus, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction for a projectile when we plug into the kinematic formulas.

Warning: Forgetting to include a negative sign is one of the most common sources of error when using kinematic formulas.

How do you select and use a kinematic formula?

We choose the kinematic formula that includes both the unknown variable we're looking for and three of the kinematic variables we already know. This way, we can solve for the unknown we want to find, which will be the only unknown in the formula.

For instance, say we knew a book on the ground was kicked forward with an initial velocity of $v_0=5\text{ m/s}$v, start subscript, 0, end subscript, equals, 5, start text, space, m, slash, s, end text, after which it took a time interval $t=3\text{ s}$t, equals, 3, start text, space, s, end text for the book to slide a displacement of $\Delta x=8\text{ m}$delta, x, equals, 8, start text, space, m, end text. We could use the kinematic formula $\Delta x=v_0 t+\dfrac{1}{2}at^2$delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared to algebraically solve for the unknown acceleration $a$a of the book—assuming the acceleration was constant—since we know every other variable in the formula besides $a$a—$\Delta x, v_0, t$delta, x, comma, v, start subscript, 0, end subscript, comma, t.

Problem solving tip: Note that each kinematic formula is missing one of the five kinematic variables—$\Delta x, t, v_0, v, a$delta, x, comma, t, comma, v, start subscript, 0, end subscript, comma, v, comma, a.

$1. \quad v=v_0+at \quad \text{(This formula is missing $\Delta x$.)}$1, point, v, equals, v, start subscript, 0, end subscript, plus, a, t, start text, left parenthesis, T, h, i, s, space, f, o, r, m, u, l, a, space, i, s, space, m, i, s, s, i, n, g, space, delta, x, point, right parenthesis, end text

$2. \quad {\Delta x}=(\dfrac{v+v_0}{2})t\quad \text{(This formula is missing $a$.)}$2, point, delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t, start text, left parenthesis, T, h, i, s, space, f, o, r, m, u, l, a, space, i, s, space, m, i, s, s, i, n, g, space, a, point, right parenthesis, end text

$3. \quad \Delta x=v_0 t+\dfrac{1}{2}at^2\quad \text{(This formula is missing $v$.)}$3, point, delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared, start text, left parenthesis, T, h, i, s, space, f, o, r, m, u, l, a, space, i, s, space, m, i, s, s, i, n, g, space, v, point, right parenthesis, end text

$4. \quad v^2=v_0^2+2a\Delta x\quad \text{(This formula is missing $t$.)}$4, point, v, squared, equals, v, start subscript, 0, end subscript, squared, plus, 2, a, delta, x, start text, left parenthesis, T, h, i, s, space, f, o, r, m, u, l, a, space, i, s, space, m, i, s, s, i, n, g, space, t, point, right parenthesis, end text

To choose the kinematic formula that's right for your problem, figure out which variable you are not given and not asked to find. For example, in the problem given above, the final velocity $v$v of the book was neither given nor asked for, so we should choose a formula that does not include $v$v at all. The kinematic formula $\Delta x=v_0 t+\dfrac{1}{2}at^2$delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared is missing $v$v, so it's the right choice in this case to solve for the acceleration $a$a.

How do you derive the first kinematic formula, $v=v_0+at$v, equals, v, start subscript, 0, end subscript, plus, a, t ?

This kinematic formula is probably the easiest to derive since it is really just a rearranged version of the definition of acceleration. We can start with the definition of acceleration,

$a=\dfrac{\Delta v}{\Delta t}$a, equals, start fraction, delta, v, divided by, delta, t, end fraction $\quad$

Now we can replace $\Delta v$delta, v with the definition of change in velocity $v-v_0$v, minus, v, start subscript, 0, end subscript.

$a=\dfrac{v-v_0}{\Delta t}$a, equals, start fraction, v, minus, v, start subscript, 0, end subscript, divided by, delta, t, end fraction

Finally if we just solve for $v$v we get

$v=v_0+a\Delta t$v, equals, v, start subscript, 0, end subscript, plus, a, delta, t

And if we agree to just use $t$t for $\Delta t$delta, t, this becomes the first kinematic formula.

$\Large v=v_0+at$v, equals, v, start subscript, 0, end subscript, plus, a, t

How do you derive the second kinematic formula, ${\Delta x}=(\dfrac{v+v_0}{2})t$delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t?

A cool way to visually derive this kinematic formula is by considering the velocity graph for an object with constant acceleration—in other words, a constant slope—and starts with initial velocity $v_0$v, start subscript, 0, end subscript as seen in the graph below.

$v~(m/s)$$t~(s)$$t$$v_0$$v$

The area under any velocity graph gives the displacement $\Delta x$delta, x. So, the area under this velocity graph will be the displacement $\Delta x$delta, x of the object.

$\Delta x=\text{ total area}$delta, x, equals, start text, space, t, o, t, a, l, space, a, r, e, a, end text

We can conveniently break this area into a blue rectangle and a red triangle as seen in the graph above.

The height of the blue rectangle is $v_0$v, start subscript, 0, end subscript and the width is $t$t, so the area of the blue rectangle is $v_0t$v, start subscript, 0, end subscript, t.
The base of the red triangle is $t$t and the height is $v-v_0$v, minus, v, start subscript, 0, end subscript, so the area of the red triangle is $\dfrac{1}{2}t(v-v_0)$start fraction, 1, divided by, 2, end fraction, t, left parenthesis, v, minus, v, start subscript, 0, end subscript, right parenthesis.

The total area will be the sum of the areas of the blue rectangle and the red triangle.

$\Delta x=v_0t+\dfrac{1}{2}t(v-v_0)$delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, t, left parenthesis, v, minus, v, start subscript, 0, end subscript, right parenthesis

If we distribute the factor of $\dfrac{1}{2}t$start fraction, 1, divided by, 2, end fraction, t we get

$\Delta x=v_0t+\dfrac{1}{2}vt-\dfrac{1}{2}v_0t$delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, v, t, minus, start fraction, 1, divided by, 2, end fraction, v, start subscript, 0, end subscript, t

We can simplify by combining the $v_0$v, start subscript, 0, end subscript terms to get

$\Delta x=\dfrac{1}{2}vt+\dfrac{1}{2}v_0t$delta, x, equals, start fraction, 1, divided by, 2, end fraction, v, t, plus, start fraction, 1, divided by, 2, end fraction, v, start subscript, 0, end subscript, t

And finally we can rewrite the right hand side to get the second kinematic formula.

$\Large \Delta x=(\dfrac{v+v_0}{2})t$delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t

This formula is interesting since if you divide both sides by $t$t, you get $\dfrac{\Delta x}{t}=(\dfrac{v+v_0}{2})$start fraction, delta, x, divided by, t, end fraction, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis. This shows that the average velocity $\dfrac{\Delta x}{t}$start fraction, delta, x, divided by, t, end fraction equals the average of the final and initial velocities $\dfrac{v+v_0}{2}$start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction. However, this is only true assuming the acceleration is constant since we derived this formula from a velocity graph with constant slope/acceleration.

How do you derive the third kinematic formula, $\Delta x=v_0 t+\dfrac{1}{2}at^2$delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared?

There are a couple ways to derive the equation $\Delta x=v_0 t+\dfrac{1}{2}at^2$delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared. There's a cool geometric derivation and a less exciting plugging-and-chugging derivation. We'll do the cool geometric derivation first.

Consider an object that starts with a velocity $v_0$v, start subscript, 0, end subscript and maintains constant acceleration to a final velocity of $v$v as seen in the graph below.

$v~(m/s)$$t~(s)$$t$$v_0$$v$

Since the area under a velocity graph gives the displacement $\Delta x$delta, x, each term on the right hand side of the formula $\Delta x=v_0 t+\dfrac{1}{2}at^2$delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared represents an area in the graph above.

The term $v_0t$v, start subscript, 0, end subscript, t represents the area of the blue rectangle since $A_{rectangle}=hw$A, start subscript, r, e, c, t, a, n, g, l, e, end subscript, equals, h, w.

The term $\dfrac{1}{2}at^2$start fraction, 1, divided by, 2, end fraction, a, t, squared represents the area of the red triangle since $A_{triangle}=\dfrac{1}{2}bh$A, start subscript, t, r, i, a, n, g, l, e, end subscript, equals, start fraction, 1, divided by, 2, end fraction, b, h. 

That's it. The formula $\Delta x=v_0 t+\dfrac{1}{2}at^2$delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared has to be true since the displacement must be given by the total area under the curve. We did assume the velocity graph was a nice diagonal line so that we could use the triangle formula, so this kinematic formula—like all the rest of the kinematic formulas—is only true under the assumption that the acceleration is constant.

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Here's the alternative plugging-and-chugging derivation. The third kinematic formula can be derived by plugging in the first kinematic formula, $v=v_0+at$v, equals, v, start subscript, 0, end subscript, plus, a, t, into the second kinematic formula, $\dfrac{\Delta x}{t}=\dfrac{v+v_0}{2}$start fraction, delta, x, divided by, t, end fraction, equals, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction.

If we start with second kinematic formula

$\dfrac{\Delta x}{t}=\dfrac{v+v_0}{2}$start fraction, delta, x, divided by, t, end fraction, equals, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction

and we use $v=v_0+at$v, equals, v, start subscript, 0, end subscript, plus, a, t to plug in for $v$v, we get

$\dfrac{\Delta x}{t}=\dfrac{(v_0+at)+v_0}{2}$start fraction, delta, x, divided by, t, end fraction, equals, start fraction, left parenthesis, v, start subscript, 0, end subscript, plus, a, t, right parenthesis, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction

We can expand the right hand side and get

$\dfrac{\Delta x}{t}=\dfrac{v_0}{2}+\dfrac{at}{2}+\dfrac{v_0}{2}$start fraction, delta, x, divided by, t, end fraction, equals, start fraction, v, start subscript, 0, end subscript, divided by, 2, end fraction, plus, start fraction, a, t, divided by, 2, end fraction, plus, start fraction, v, start subscript, 0, end subscript, divided by, 2, end fraction

Combining the $\dfrac{v_0}{2}$start fraction, v, start subscript, 0, end subscript, divided by, 2, end fraction terms on the right hand side gives us

$\dfrac{\Delta x}{t}=v_0+\dfrac{at}{2}$start fraction, delta, x, divided by, t, end fraction, equals, v, start subscript, 0, end subscript, plus, start fraction, a, t, divided by, 2, end fraction

And finally multiplying both sides by the time $t$t gives us the third kinematic formula.

$\Large \Delta x=v_0 t+\dfrac{1}{2}at^2$delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared

Again, we used other kinematic formulas, which have a requirement of constant acceleration, so this third kinematic formula is also only true under the assumption that the acceleration is constant.

How do you derive the fourth kinematic formula, $v^2=v_0^2+2a\Delta x$v, squared, equals, v, start subscript, 0, end subscript, squared, plus, 2, a, delta, x?

To derive the fourth kinematic formula, we'll start with the second kinematic formula:

${\Delta x}=(\dfrac{v+v_0}{2})t$delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t

We want to eliminate the time $t$t from this formula. To do this, we'll solve the first kinematic formula, $v=v_0+at$v, equals, v, start subscript, 0, end subscript, plus, a, t, for time to get $t=\dfrac{v-v_0}{a}$t, equals, start fraction, v, minus, v, start subscript, 0, end subscript, divided by, a, end fraction. If we plug this expression for time $t$t into the second kinematic formula we'll get

${\Delta x}=(\dfrac{v+v_0}{2})(\dfrac{v-v_0}{a})$delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, left parenthesis, start fraction, v, minus, v, start subscript, 0, end subscript, divided by, a, end fraction, right parenthesis

Multiplying the fractions on the right hand side gives

${\Delta x}=(\dfrac{v^2-v_0^2}{2a})$delta, x, equals, left parenthesis, start fraction, v, squared, minus, v, start subscript, 0, end subscript, squared, divided by, 2, a, end fraction, right parenthesis

And now solving for $v^2$v, squared we get the fourth kinematic formula.

$\Large v^2=v_0^2+2a\Delta x$v, squared, equals, v, start subscript, 0, end subscript, squared, plus, 2, a, delta, x

What's confusing about the kinematic formulas?

People often forget that the kinematic formulas are only true assuming the acceleration is constant during the time interval considered.

Sometimes a known variable will not be explicitly given in a problem, but rather implied with codewords. For instance, "starts from rest" means $v_0=0$v, start subscript, 0, end subscript, equals, 0, "dropped" often means $v_0=0$v, start subscript, 0, end subscript, equals, 0, and "comes to a stop" means $v=0$v, equals, 0. Also, the magnitude of the acceleration due to gravity on all freely flying projectiles is assumed to be $g=9.81\dfrac{\text{m}}{\text{s}^2}$g, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, so this acceleration will usually not be given explicitly in a problem but will just be implied for a freely flying object.

People forget that all the kinematic variables—$\Delta x, v_o, v, a$delta, x, comma, v, start subscript, o, end subscript, comma, v, comma, a—except for $t$t can be negative. A missing negative sign is a very common source of error. If upward is assumed to be positive, then the acceleration due to gravity for a freely flying object must be negative: $a_g=-9.81\dfrac{\text{m}}{\text{s}^2}$a, start subscript, g, end subscript, equals, minus, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction.

The third kinematic formula, $\Delta x=v_0 t+\dfrac{1}{2}at^2$delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared, might require the use of the quadratic formula, see solved example 3 below.

People forget that even though you can choose any time interval during the constant acceleration, the kinematic variables you plug into a kinematic formula must be consistent with that time interval. In other words, the initial velocity $v_0$v, start subscript, 0, end subscript has to be the velocity of the object at the initial position and start of the time interval $t$t. Similarly, the final velocity $v$v must be the velocity at the final position and end of the time interval $t$t being analyzed.

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