How do you find velocity if time is only given ???
Velocity is distance travelled per unit time in a given direction.
So if only time is given, you cannot find the velocity. Try to look for hidden information which can be translated into distance.
The same is true for direction, which is usually understood for rectilinear motion.
Well the problem gives me a graph with Vx on the (Y axis) and time on the (X axis). but only says Find the velocity at t = 4.0 s and t = 7.0 s... would i look on the graph and see where they hit the line??
Exactly!
Go to where x-axis corresponds to t=4.0 s. and draw a vertical line through this point. At the intersection of this vertical line and the graph, find the value of y, which corresponds to the velocity.
The graph is a summary of the different velocities at different times. It is the "hidden" information that gives you all the rest.
so I got t=4s so Vx (cm/s) does it matter if it is in cm/s does it need to be in m/s??? but i got a velocity of 3 and for t=7.0s i got a velocity of -1. do I need to convert those to m/s?
Then to find acceleration I do V2x - V1x / t2-t1 right?
If everything else in your problem is in cm and seconds, you can leave the units. One good way to avoid errors is to make sure you write the units when you do your calculations. This will help you check if you have suitable units.
V2x - V1x / t2-t1 is correct if you have the parentheses at the right place, AND if the units are consistent.
I prefer to see parentheses as follows to avoid ambiguity:
acceleration=(V2x - V1x) / (t2-t1)
ok now here is the tricky part they threw in the problem they want me to find the acceleration for the two times above the 4 and 7 and for now t = 6.0? if there are only 2 spots in the acceleration equation where does my third number go?
You will have to either calculate the velocity at t=6 s and calculate the acceleration as before using:
acceleration=(V2x - V1x) / (t2-t1)
You can also read off the value of velocity on the graph using the vertical line through t=6 s as I explained before.
Is the graph between times 4s and 7s a straight line? If it is, then the acceleration is uniform, which means that it is the same everywhere between 4 s and 7s (including t=6 s). For your benefit, go through the calculations anyway.