# How to sketch a "2nd grade s-t" from a "linear v-t"

1. Oct 15, 2015

### Optikspik

1. The problem statement, all variables and given/known data (with s i mean what many people know as "x" also known as "distance"
IF this v-t diagram was given (Just made up in my head) , how would i sketch a s-t diagram?

Since the s-t diagram would be a second grade polynomia, for instance between t=0 and t=5 the area under the curve on the v-t would be 1/2*5*10=25 (in the unit meter).

So between t=0 and t=5 s would go from 0 to 25 meters, but how can i draw it "perfectly" without just "sketching?

Anyone have a tip?

For instance:

Can i make the s-t skiss "perfect by doing an a-t diagram first or? How would u attempt an solution?

Or if no numbers were given, would from t=0 to t=5 look like this , or would the (dy/dx) be more and then decrease? or is it like here that the differentiate (dy/dx) is increasing all the time on s-t between t=0 and t=5

2. Oct 15, 2015

### slider142

The velocity function v(t) is, by definition, the derivative function of the position function s(t). If you can find the exact form of the velocity function, you can then find a form of the position function by finding an antiderivative. The many possible antiderivatives differ by a single additive constant, which requires knowledge of at least one exact position at an exact time.
In particular, assuming the section of the graph between (0,0) and (5, 10) is a straight line, we can calculate its slope as $\frac{10}{5} = 2$. As the y-intercept is 0, we know that this part of the velocity function may be calculated with the function v(t) = 2t. Any position function s(t) whose derivative is 2t could then be a potential position function that would have that particular section of the graph as ithe graph of its velocity function.
If, hypothetically, you did not know of any function you could graph which has 2t as its derivative function, then you would start at some known position, such as s(0) = 0, and attach a line of slope 2(0) = 0 for a small distance, like 1 unit. Then, at (1, 0), attach a tangent line of slope 2t = 2(1) = 2 for another unit, so that it ends at (2, 2). And so on. In reality, you may want to program a computer to do this, and use much smaller intervals than 1 unit. The smaller the interval you use, the closer the graph will be to the actual solution. This method of sketching the graph of an unknown curve with a known derivative is known as Euler's method.

3. Oct 15, 2015

### RUber

From t= 0 to 5, it looks like v(t)=2t.
So, you would plot s(t) = t^2 over that range.
Then it appears v(t) is constant for 4 seconds at 10, so s is increasing linearly at 10units/second.
A drop in v(t) will be a slowdown but still increasing.
It looks like you have a good start.