Full transcript
0:00if we have a car traveling at a constant
0:02speed we can simply find at speed by
0:04doing distance over time in another
0:07situation this course starts with a
0:09certain speed and then it speeds up this
0:11equation wouldn't work now because it's
0:14only applicable to an object traveling
0:16with a constant speed we need to use
0:18some other equations of motion let's
0:21define the initial velocity to be u the
0:23final velocity V and distance between
0:26its starting point and plan a position
0:28is two displacements s a is the
0:31acceleration and T is time these
0:34variables are related by Newton's
0:36equations of motion and there are four
0:38of them number one the final velocity is
0:41equal to the initial velocity plus the
0:44acceleration times time this makes a lot
0:46of sense because this here is to gain in
0:49velocity another one is V squared is U
0:52squared plus 2ei s third equation s is
0:56UT plus half a T squared
0:58and lastly SS u plus B divided by two
1:01times time we can also easily explain
1:04this equation because this chunk here u
1:06plus V divided by two is the average
1:08velocity as put these aside and look at
1:11an example a car is initially at rest
1:13and after five seconds it has
1:15accelerated to twelve meters per second
1:18we have to find the acceleration I like
1:20to use what's called a tool box method
1:22we write down all the variables that we
1:24know and these will be our tools we
1:26don't know the displacement so I'll
1:28leave it blank the car was initially at
1:30rest so the initial velocity is zero
1:33final velocity is 12 meters per second
1:36acceleration is what we want to find so
1:38I'll put a question mark next to it t is
1:41five seconds so now I need to select an
1:44equation that will only involve UV a and
1:47t I can't use any equations with
1:50displacement s involved so it seems like
1:52it can use the first equation but I need
1:54to rearrange it first
1:56making acceleration the subject to bring
1:59you to the other side I subtract and to
2:01bring a T to the other side I divide
2:04substituting numbers and from the
2:05toolbox we have 12 minus 0 divided by 5
2:09and that is equal to two point four
2:11meters per second squared for our
2:13acceleration let's look at another
2:15example and drop a ball from the top of
2:17the cliff and after 20 seconds it hits
2:20the bottom and want to find the heights
2:22of the cliff
2:23so essentially we want to find the
2:25displacement of the ball because that is
2:27equal to the heights of the clip again
2:29using the tool box method writes out SUV
2:3280 soo that is what I'm trying to find
2:35so here's the question mark the ball was
2:37initially in my hand at rest so it had a
2:40velocity of 0 I don't know the final
2:42velocity of the ball but I know
2:44acceleration due to gravity is always
2:469.8 meters per second squared on earth
2:49and a question tells us that time is 20
2:52seconds I can use the third equation s
2:54is UT plus half ay T squared this
2:57becomes 0 because U is 0 so s is 1/2
3:01times 9.8 times 20 squared giving us a
3:05height of 1960 meters here's a summary
3:08of what to do when using su that's
3:10equations number 1 write down all the
3:13known variables given by question number
3:162 identify the unknown this is what the
3:19question wants us to calculate number 3
3:21selects the equation that only involves
3:23the variables mentioned in a question
3:25and before if needed want to rearrange
3:28equation to make the unknown variable
3:31the subject and finally substitute the
3:34numbers in for your calculation