Velocity, position, acceleration graph help

[Physicshelpneeded]

im new to this site...so i don't kno whow this works...ANYWAYS...

so you guys know : I am 16, taking a college course at mt sac (physics 1)...and I am kind of struggling because my math isn't as high as it nees to be for the coruse and because i don't understand soem concepts.

im having trouble with a velocity/time graph problem... the question asks me to determine the distance moved by the object in the time interval from t=0 to t=0.4 (since u guys can't see the graph, i will tell u...the velocity moves from 3.4 to 3.9 in 4 seconds) anyone know how to do this?

the second part to this problem is What additional information do you need to determine the position of the object at t=0.4 s? explain.

CAN ANYONE HELP ME WITH THIS? Thanks guys! (and if you are a girl, girls)

 

[Blistering Peanut]

There are two ways of doing this but you want to do it the graph way. The most important thing to remember is that the area under a velocity/time graph is equal to the distance travelled. So isolate the area under the graph from said interval and divide it up into simple shapes - squares/triangles. It's just very easy geometry from there.

For the second half I'm not sure what they mean by that.

 

[Physicshelpneeded]

thanks for the advice! :) it helped a littel bit...but the problem i have is that the graph is curved...and makes a triangle and an almost-full rectangle...but not an actual one...so how would i find the area of the weird shape? is there a secret way or formula?

and can anyon eelse help me with part 2?


thanx

 

[rayjohn01]

I think Mr peanut already said this , if you must do it graphically ( and the shape is not a rectangle - then you divide the shape into small but measureable sections. Typically use a given small section of time say 0.1 seconds and divide the x-axis ( time ) into a number of the sections.In each section measure the y value ( velocity) -- the strip area is thus v*dt . Add up all the sections to get the total area the result is the distance.The accuracy is dependant on the size of dt chosen.
Do you have a typo in the question you said the velocity changes from 3.4 to 3.9 in '4' seconds . did you mean 0.4.

 

[Physicshelpneeded]

yes, i meant 0.4...thanx for ur help...does anyone else know anything about this particular question?


and can ANYONE answer part 2?

 

[arildno]

Part 2:
Assuming you know the distance traveled in the time interval, you'll also need the initial position in order to determine the final position
(I assumed that during the time interval the particle moved in a straight line from the initial position)