Leah here from leah4sci.com/mcat and in this translational motion video I'll take you through
the concept of Acceleration.
You can find my entire video series on Translational Motion by visiting my website http://leah4sci.com/MCATPhysics
Unlike the topic of Velocity, Acceleration can be a little more difficult to understand
intuitively.
Acceleration is a Vector quantity defined as the change in velocity over the change
of time.
In other words, we're looking at how quickly something can increase or decrease in speed
usually at a consistent rate.
Think of this intuitively if you are in a car trying to get off to the highway, your
velocity before the highway is slow.
On the highway is fast.
The rate which will you increase will be your acceleration.
Sometimes it take a while to get to that fast velocity and sometimes you hit the accelerator
and you increase your acceleration really quickly to get to that maximum speed.
Let's try a simple example: A cyclist increases his velocity from rest
to 4.5 m/s in a span of 5 seconds.
Find his acceleration.
The formula for acceleration is the change of velocity over the change in time or simply
change in velocity over time.
If we break this apart, delta v is really V final minus V initial, you'll also see it
written as V with a small zero which stands for V knot or the initial velocity and we
divide that by t.
Plugging in the numbers, we have 4.5 meters per second as the velocity and 5 seconds is
the time.
Always mark your units and cancel units to make sure you're on the right path.
We're looking at the equation of 4.5 divided by 5 which involves a decimal and a fraction,
but you have to do that without a calculator.
Don't let this confuse you, we'll use the multiply the top and bottom by ten trick or
move over the decimal so we'll move the decimal to the right, the imaginary decimal to the
right.
Now we have 45 divided by 50 which you can think of as 45 over 5 times 10.
This makes it so much easier.
Both 45 and 5 can be divided by 5.
Forty five by five is nine, five by five is one and that gives us a new equation of 9
over 1 times 10 which is 9 over 10.
Now we'll use the trick of dividing by ten.
10 has one zero so we move the decimal back one space and the answer is 0.9.
If you're not comfortable with the Math I used here, go back to my website http://leah4sci.com/MCATMath
where I have an entire video series showing you all these MCAT Math without a calculator
trick.
Now let's take a look at the units.
Velocity was given in meters per second and time was given in per second because it's
in the denominator.
That means our units are now is meters over seconds times seconds which is really meters
per second squared (m/s^2) the unit for acceleration giving us a final answer of zero point nine
meters per second squared (0.9m/s^2).
That was a pretty straightforward example, now let's play with the equation a little
bit so that you're comfortable solving Acceleration questions no matter how it's presented.
The equation for acceleration is given as the change in velocity over the change in
time.
But what if you're asked to solve for something like the Final Velocity?
To solve for Velocity, the first thing we want to do is to isolate it so we'll multiply
both sides by t, this cancels it out on the right giving me a new version of the equation,
delta V is equals to a times t.
Delta V is V final minus V initial equals "at".
And now let's say we want V final let's move V initial to the other side.
It'll be plus V initial, plus V initial.
Let's rewrite it into the form which you're probably more familiar with, V final is equal
to V initial plus at (Vf =V i+ at).
This is the form that I initially studied when I learned this in Physics, this is also
the form that you'll see on my Kinematics cheat sheet on my website at http://leah4sci.com/MCATPhysics.
Keeping in mind, many times your Acceleration question will have an initial Velocity of
zero just cancel that out, giving you an even simpler form of V final is equal to at or
simply V is equal to at.
Now let's say you're asked for Time.
To solve for time in the denominator, the first thing we wanna do is move it to numerator,
we'll multiply both sides to cancel it out this gives us delta V is equal to at.
Now we wanna isolate T so we'll divide both sides by the acceleration giving us a new
version of the same equation t is equals to delta v over a.
I worked through these because I don't want to fall into the trap of memorizing every
single form.
Instead I want you to understand the formula, understand how to move the unit around so
that no matter how it's asked, you'll know how to tackle the question.
Let's try another example: Our cyclist from above is coasting at 7.5
meters per second when he hits the breaks.
How long does it take him to reach a complete stop if he slows with an acceleration of -2.5
m/s^2?
The first thing to notice is that we have a negative number for acceleration.
This means that our final velocity is less than our initial velocity giving us the negative
sign and the negative is simply a direction.
Instead of accelerating to go faster he is decelerating to go slower or simply acquiring
a negative acceleration.
This seems like a simple equation, except that we�re asked for time instead of acceleration
or velocity.
So we�ll manipulate the equation as we saw earlier.
We�ll use the version t is equal to delta V over the acceleration in this case delta
v is V final minus V initial.
V final is zero, V initial is seven point five meters per second, Acceleration is negative
two point five meters per second.
Time cannot be a negative unit and we verify that here because the negative change in velocity
and the negative acceleration cancel out.
Leaving us with the calculation of seven point five over two point five.
Once again we�re dealing with decimals and fractions which is confusing so we�ll use
the trick of moving the decimal over to the right for the numerator and denominator, this
gives us an easier version of the equation seventy five over twenty five.
Both numbers easily divide by twenty five.
Twenty five into Seventy is three, twenty divided by twenty five is one.
Three over one is simply three.
Now let�s take a look at our units.
We have meters per second divided by meters per second squared.
Meters cancels meters and seconds cancels one of the seconds in second square which
is seconds times second leaving us with just seconds as our answer because one over second
is second for a final answer of three seconds (3s).
Be sure to join me in the next video where I tackle a more complex form of acceleration
as it comes up in Gravity and Free fall questions.
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