1. Jan 4, 2010

### james1232

im a grade 11 student and i got a question that i dont understand, here it is. 32) a) derive an equation for the change in acceleration by using a graph analysis technique on a a-t (acceleration vs time) graph. (note. The name given tothe change in acceleration per unit time is called a jerk)

b) write an equation for the area under the graph . what does this represent.

so I thought that the equation to calculate acceleration in a = v/delta t but i dont have any clue what the equation is for the change in acceleration, i figure maybe if they were two straight lines you could just do a2 - a1 = change in acceleration, but on a a-t graph how would that work, because it would have to be two straight lines, not a diagonal line... if any one has any ideas that would be great.

2. Jan 4, 2010

### jgens

You might clear your confusion up by clarifying your terminology. For example, the jerk is not defined as the change in acceleration but rather the change in acceleration per unit time. Since you're presumably in calculus you should know that velocity (which is the change in position per unit time) is given by $dx/dt$ and that acceleration (which is the change in velocity per unit time) is given by $dv/dt$. What does this suggest about the jerk? How does the jerk relate to an acceleration v. time graph?

3. Jan 4, 2010

### james1232

thank you for the clarification, i did understand that a jerk was change in acceleration per unit time, guess i forgot to put that down. oh well, im not sure how it relates to the graph, we were not given any graph at all, just the question, so i presume that we were to jsut rough our own, but im not sure, but i do see what you mean that i need to think of it as a change in acceleration per unit time, but its still kind of confusing, thank you for clearing a bit of my confusion up.

4. Jan 4, 2010

### jgens

Well, if I were to give you a graph that charted the displacement of an object over time, could you tell me how you might find the instantaneous velocity?

5. Jan 4, 2010

### james1232

oh yea that easy, if its a curve, which is the only graph where in this case there would be a different instantaneous velocity, that easy though, i wold carefully take a ruler, find the spot on the line where there is two eqal spaces from the line to the ruler on both sides, then draw the line, and find two points on that line that meet up nicely with the corners of the graph paper sqaures, and then use those to find the slope, and that would be the instantaneous velocity

6. Jan 4, 2010

### jgens

So . . . for a curve tracing displacement v. time, the instantaneous velocity ($dx/dt[/tex]) is given by the slope of the tangent line. Now suppose that we have a curve tracing acceleration v. time. How might I find [itex]da/dt$ or the instantaneous rate of change of acceleration with respect to time?

7. Jan 4, 2010

### james1232

same thing, but we needed an equation to find the difference, so are you saying to find two different instantaneous intantaneous rates of change of acceleration ???

8. Jan 4, 2010

### jgens

Nope. Since you say you aren't given an equation or curve, let $a$ be the acceleration function, then $a(t)$ gives the instantaneous acceleration at time $t$. Now because you're presumably in calculus, given some function $a$ defined by $a(t)$, how do you find the instantaneous rate of change of $a$?

9. Jan 4, 2010

### james1232

hmm im embarassd to say i dont think i understand what you mean by a function... haha sorry, im trying to think of what you mean so just let a represent acceleration?? okay, and a(t) so a multiplied by time? wouldnt you just input the function given, multiply it by the point in time where you want to find the change in velocity so maybe from like 0s to 5 seconds, you would find the two then subtract the second one from the first one?

10. Jan 4, 2010

### jgens

Before moving further forward, you are in calculus, correct? Assuming you are in calculus, are you familiar with the concept of a derivative?

11. Jan 4, 2010

### james1232

um no and im in physics, not positive its exactly calculus, but i know he said it contained calculus, it was sort of a review thing i can check what unit its from hold on

12. Jan 4, 2010

### james1232

i assume it had to do with uniform motion i guess, we werent taught anything about acceleration vs time graphs, it was just a question to confuse us and make us miserable and get us thinkin, makin connections but i seem to make things too complicated, so i apologize for any inconveniance, i didnt really know which other forum to post this on, he said it contained calculus pieces, but i wasnt exactly sure what genre it would fall under my mistake,

13. Jan 4, 2010

### james1232

like uniform motion graphs and the equations we got from the graphs seem sort of similar to this so thats my only reasoning for saying i assume its uniform motion

14. Jan 4, 2010

### jgens

Alright, that might explain why you don't know what a function is. Even without knowledge of calculus, you should be able to intuitively understand the first problem (writing out an equation like $j(t) = a'(t)$ would be a bit difficult though). For the second problem, you would find that the area under the curve is the velocity and that it's given by

$$\int_{t_1}^{t_2} a(t)\mathrm{d}t = v(t)$$

which if you're given a specific curve may be evaluated directly.

15. Jan 4, 2010

### james1232

AHAH!! thats what i wrote down when i handed it in so i got that right, but yea okay thanks alot man appretiate that,
but one thing i dont understand, and maybe im just stupid but ive seen the symbol before on the forums and ive never been told what it is , the like that look kind of like a square root sign with t1 at the bottom and t2 at the top, what is that called and what does it mean, is it jsut like delta t?? i have no clue

16. Jan 4, 2010

### jgens

Presumably you're talking about the integral sign: $$\int$$

Integration is an operation that may be used to evaluate the area bounded by a curve. As it turns out, integration is closely tied with anti-differentiation but this isn't the way integrals are defined. There are several ways of defining the integral, but unless you're familiar with limits or supremums and infimums, I won't provide you with a formal definition. If you're interested in the general concepts behind them, google "Riemann Sum" and you should find some relevent articles.

17. Jan 5, 2010