The graph at the right is a positive vs. Time graph for the motion of an object moving along a horizontal (x) axis For each of the 8 1-second time intervals, describe the displacement, instantaneous velocity (+/- and whether the velocity is increasing or decreasing) and the instantaneous acceleration of the object (+/-) for that individual 1-second time interval. Note that these descriptions can change inside of each of the time intervals ( i.e. during a single interval, the velocity may start off constant, but could change in the middle of that 1-second interval).
Since I can't get a visual represenation to send, I'll describe it so you can draw it to see:
Position Vs. Time graph (8x8) grid!
Position (y axis) ... x(m)
Starts at 0, count by 20's to reach 80 at the very top line on the 8x8 grid.
Time (x axis) ... t(s)
Starts at 0, count by 1 s all the way to 8 s (which should be on the very last line of the 8x8 grid.
PLEASE TELL ME IF YOU CAN'T VISUALLY DRAW THE GRAPH SO I CAN EXPLAIN IT BETTER OR DESCRIBE IT BETTER. I TRIED MY BEST TO DESCRIBE THE GRAPH FOR YOU TO HELP ME. Thankyou!
Now here are the points on the graph:
1.) From 0-1 second, the line is constant at 43m.
2.) From 1-2 seconds, the line is a positive slight under curve from 43m to point (2,60).
3.) From 2-3 seconds, the point goes from (2s, 60m) to (3s, 70m) at a positive slight over curve(slowing down but still positive).
4.) From 3-4 seconds, the line is constant from (3,70) to (4,70).
5.) From 4-5 seconds, the line is slightly decreasing from point (4,70) to points (5,68).
6.) From 5-6 seconds, starts from point (5,68) and decreases at a straight diagonal line to (6s, 2m).
7.) From 6-7 seconds, starting from point (6s, 2m), going to points (7,10) at a diagonal line but half the box, curved.
8.) From 7-8 seconds, the line is constant and ends at (8s, 10m)
HOPEFULLY, YOU CAN UNDERSTAND AND DRAW THAT. I TRIED TO DESCRIBE THE POINTS AND LINES OF WHERE IT IS GOING.
________________________________________
Now fill in the chart of the intervals describing its specific VELOCITY at that time during how many meters, and the ACCELERATION!
* I had to describe the position of each interval. I got all those right but the velocity column and acceleration column, I really need help.
______________________________________ |Describe Velocity| & separately |Describe Acc.| for each intervals!
INTERVALS (seconds)
1s
2s
3s
4s
5s
6s
7s
8s
time is ending at t time
eg = 1 sec is ending at 1, first second
v is average velocity in inteval
(final-original)
a is acceleration change in speed/1 second
CHECK MY WORK
___ v a
1s 00 00
2s 17 10 v=(60-43) a =(v3-0)=10
3s 10 -17
4s 00 -2
5s -2 (2m-66)
6s (2m-68) -58
7s (10-2m) (2m -10)
8s 00 (10-2m)
everything is divided by one second
for velocity I take position
at n minus position at (n-1)
for acceleration I take velocity during n+1 - velocity during n
By the way I am assuming that your curve is continuous and smooth.
If there is an instantaneous change in slope (velocity), for example at 1 second, then the acceleration is infinite. a = (V2-V1)/0
CHECK MY WORK
___ v a
1s 00 00
2s 17 10 v=(60-43) a =(v3-0)=10
3s 10 -17
4s 00 -2
5s -2 (2m-66)
6s (2m-68) -58
7s (10-2m) (2m -10)
8s 00 (10-2m)
I got lost with the numbers. Could you please explain in words.. I am still lost. Hopefully I get a reply before 11 am so I can process the info before 12. Thanks Mr.Damon
3 answers
Perhaps this will help you.http://dev.physicslab.org/Document.aspx?doctype=5&filename=Kinematics_ConstantVelocityPositionTimeGraphs2.xml
velocity=Vn = (Xn - Xn-1)/1 second
acceleration = (Vn-Vn-1)/1 second