Today, we wrapped up unit three and moved on to unit four: 2D kinematics. A "solve for one variable" type of kinematic equation quickly turned into a more complicated multi-step process. This unit made understanding physics much harder.
We learned to set up a 2D kinematics equation with a T graph. (Shown below.) Some of the most important things to remember are to separate x axis things from y axis things (vegas rule) and time is a value of y (aYer).
If I were to dive off a diving board 5 meters above the ground, going forward at 1.5 m/s, how far would I be from the bottom of the diving board to the place I landed?
___x___|___y___
?m d 5m (height of diving board)
0 m/s/s a -9m/s/s(gravity)
t
1.5 m/s v (given)
1.5 m/s vo 0 m/s (given)
Now I have to solve for time using d=1/2at^2+Vot
5=1/2(-9.8)(t^2)+(0)(t)
1.02=t^2
1.01s = t
This means it took 1.01 seconds for me to hit the water.
The next step is to solve for distance of x using d=1/2at^2+Vot
d=1/2(0)(1.01)^2+(1.5)(1.01)
d=(1.5)(1.01)
d= 1.515 m
My horizontal range is 1.515 meters.
I really liked your post. It went over a lot of what we learned and showed the work along the way. Good job!
ReplyDeleteThanks Juliet! :)
ReplyDeleteGood job, but please make sure that your units in your t-chart are correct. You will need to correct your acceleration units to be m/s/s.
ReplyDeletePretty sweet that you used diving at a higher level that poolside, this is kind of like those cliff of impending death problems we did. Also nice drawing and t chart.
ReplyDeleteThanks Blake!
ReplyDelete