Kinematic Equation Problem Solving

Kinematic Equation Problem Solving
Today I'm going to present a problem which can be solved by using a kinematic equation and explain how it is done:

A penny is dropped down a wishing well and the time taken before it hits the water is 4.5 seconds. Find the distance between the top of the well and the water's surface.

To solve this problem we need to use the equation:

   where d is displacement (or in this case the distance) - the distance the penny travels from a to b or the distance displaced,
Vi is initial velocity,
t is time,
and a is acceleration.

Seen as we do not need to rearrange the equation in this example we can go ahead and put the values in.

The problem only gives us 1 value and we have to infer the rest, this is the time (4.5 seconds).
We also can tell that the initial velocity is 0m/s as we can assume that  the penny accelerates from rest.
Now for the acceleration - which is a little tricky ...

This value, simply put, is determined from acceleration due to gravity as the penny is falling - and since Earth's standard gravity is 9.8g, the acceleration of the penny is -9.8m/s^2. (the value is negative because of the direction of the penny - the penny is falling).

Now we have the values:

t = 4.5s

Vi = 0m/s
and a = -9.8m/s^2.

So 0m/s*4.5s+0.5*(-9.8m/s^2)*(4.5s)^2 can be worked out as -99.225m.

The value is negative in this case because the penny has 'displaced' the 99.225m in a downward motion and therefore the depth of the well (from the top to the surface of the water) is 99.2m.

Lets look at another example of this equation:

If a cheetah accelerates up to 27m/s in 3 seconds what is the acceleration of the cheetah and what is the distance it travelled in that time?

Again, to solve this problem the same equation is used without rearranging.

This time we are given two values but have two things to work out.

We're given the Vf (final velocity) this time which is 27m/s and we can also say again that the cheetah accelerated from rest so the Vi is once again 0m/s.
We're also given the time taken which is 4.5.

Using these two values we can work out the acceleration using the equation:

By using this we can work out that the cheetah accelerates at 9m/s^2.

Now we have all the values needed to work out the distance so:

Vi = 0m/s,
t = 3s,

and a = 9m/s^2.

So now we just need to put the values into the equation:

0m/s*3s+0.5(9m/s^2)(3s)^2 = 40.5m.

The displacement - or distance travelled by the cheetah - is 40.5 meters and the acceleration of the cheetah is 27m/s^2.

Thanks for reading, Chris.