High School How can I approach drawing v-x graphs for projectile motion problems?

[Abhishek332211]

Hello sir,
I'm a school going student, and this is my first post here. I'm having problems drawing v vs x graphs. I'm unable to understand what should be my general approach while trying to draw v-x graphs. (I've no problem drawing x-t, v-t, or a-t graphs). Let's say, for example, a ball is thrown up with some initial velocity 'u = 50m/s', it keeps going up until it's velocity becomes zero. So I put the values in 3rd eqn of motion. And this is the equation I end up with : y² = 2500 - 20x
(Here y represents velocity and x is displacement) If I were to graph this equation, what should be my general approach? Thanks a lot.

 

[FactChecker]

Method 1 (using values of t):
If you have no trouble with x-t and v-t graphs, then make a table of t, x, v, for a good set of times t. That will give you a set of values for x and v at matched times, t. Then you can plot those x-v values.

Method 2 (solving in terms of x):
From your equation v² = y² = 2500 - 20x, take the positive square root of both sides to get
v = sqrt( 2500 - 20x).
Pick a good set of x values and plot those with the corresponding values of v.

 

[sophiecentaur]

I'm having problems drawing v vs x graphs.

I guess you really mean you have problems interpreting what graphs actually do for you. You cannot believe that what you are drawing is meaningful for the x/v graph. At School, you are so often told to plot a graph but far less often are you required to get information back off a graph - but that is what graphs are for.

If you are watching an event occurring, you are aware of time moving on in equal steps (as the hand of a stop watch moves round) A x/t or v/t graph displays something in terms that you can easily appreciate. Mostly, you do your measurements or calculations for equally spaced values of t. This is not actually necessary.
A x/v graph is less tangible and you may just have to start off accepting that it is 'right' (assuming that you have drawn it correctly). As @FactChecker says, drawing up a table from all your various values of t will get the numbers down so that you can use them. When you try to plot v against x you will not get your points in a nice regular spacing. It doesn't matter. A smooth curve, drawn through those points is just as valid as drawing a smooth curve through a regularly spaced set of points. You can go to an x value and read off what the velocity will be at that place.

PS I remember I was marking a boy's homework and he had drawn a graph of results of a lab experiment. He had put all his dots along a straight diagonal line, on the corners of the 10mm graph paper squares and written the values of each variable (I can't remember what the variables were) on the X and Y axes. I asked him why he had done it that way and he said "My Mum told me to do it that way". Funny thing was that the graph actually had all the information about his experiment on it (all the data was there). It was just in a very indigestible form. You don't have that problem, Abhishek. You can draw the graph right and you can get information back off it.

 

[Abhishek332211]

I guess you really mean you have problems interpreting what graphs actually do for you. You cannot believe that what you are drawing is meaningful for the x/v graph. At School, you are so often told to plot a graph but far less often are you required to get information back off a graph - but that is what graphs are for.

If you are watching an event occurring, you are aware of time moving on in equal steps (as the hand of a stop watch moves round) A x/t or v/t graph displays something in terms that you can easily appreciate. Mostly, you do your measurements or calculations for equally spaced values of t. This is not actually necessary.
A x/v graph is less tangible and you may just have to start off accepting that it is 'right' (assuming that you have drawn it correctly). As @FactChecker says, drawing up a table from all your various values of t will get the numbers down so that you can use them. When you try to plot v against x you will not get your points in a nice regular spacing. It doesn't matter. A smooth curve, drawn through those points is just as valid as drawing a smooth curve through a regularly spaced set of points. You can go to an x value and read off what the velocity will be at that place.

PS I remember I was marking a boy's homework and he had drawn a graph of results of a lab experiment. He had put all his dots along a straight diagonal line, on the corners of the 10mm graph paper squares and written the values of each variable (I can't remember what the variables were) on the X and Y axes. I asked him why he had done it that way and he said "My Mum told me to do it that way". Funny thing was that the graph actually had all the information about his experiment on it (all the data was there). It was just in a very indigestible form. You don't have that problem, Abhishek. You can draw the graph right and you can get information back off it.

Thanks for all the help and suggestions regarding the approach that I should have while trying to draw v-x graphs. Sorry I didn't have access to internet, hence I'm responding late. And yeah, that's exactly what my problem is (was). Prior to drawing it, I had no idea what the curve will look like. I don't know how to put it. Let me put it this way - Let's say a ball is thrown straight up, and if I'm asked to draw it's x-t, v-t, or a-t graph, I can easily draw them without even making a table and working out different values of y for different values of x. Because in case of an x-t graph, the slope (i.e. velocity) will keep decreasing, and until the slope becomes zero, it's displacement will keep increasing, and then it'll have a negative slope (as it acquires a negative velocity) and then the slope will keep getting steeper and steeper. It's v-t graph is even easier to draw as the ball experiences a constant retardation. The point I'm trying to convey is, as I said earlier, I can easily draw x-t and v-t graphs without even making a table and working out different values of y for different values of x. But in case of a 'v vs x' graph, I just couldn't imagine what the 'v vs x' curve of a ball thrown straight up will look like. I had no idea whatsoever. May be, because the slope of a v-x curve has no physical meanings? I don't know, but I just couldn't figure out what the curve would look like.
I tried drawing the graph by making a table, and this is what I got. I want you to have a look at it and please let me know if I've made any mistake :)

 

[sophiecentaur]

@abhishek : First of all, let me congratulate you for drawing that graph in that orientation. There is a great temptation to draw 'up' 'upwards' and you avoided doing that.
It's OK, of course - you did the sums correctly and that should give you confidence.
When I was at school, we had no computers etc. etc. so we were taught more analytical methods for 'sketching' the graphs of functions. I'm not sure if modern day students are taught this approach. You have a formula, relating two variables and you want to know that the graph will look like. You can tell where the zero crossings are, in this case: the maximum + and - velocities will be at height = zero and you know that the speed will be a minimum at the maximum height. If the trajectory is vertical then the velocity will be zero sat maximum height and there will be a (smooth) turning value of speed around the top of the trajectory. The motion is under constant acceleration so nothing can change fast. If the trajectory is vertical, the velocity will be, assuming energy is conserved. So, without doing any calculations, you can predict the curve will have the sort of shape you have plotted. Potential energy gained at the top will be equal to the Kinetic Energy lost so mv²/2 = mgh, which tells you the maximum velocity. The shape of x/v is a parabola (from the formula) so you can sketch one, freehand, from the zero crossing on the y-axis and the zero crossing on the x axis.
I haven't mentioned "physical meaning" explicitly but Energy is clearly part of this so there is a Physical relationship between the two variables. It's just an 'unfamiliar meaning'.