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Physics 8 Online
OpenStudy (anonymous):

A car accelerates from rest on a straight road. A short time later, the car deccelerates to a stop and then returns to its original position slowing to a stop. HOW CAN ONE BEST PLOT A SKETCH OF THE POSITION -TIME GRAPH

OpenStudy (fifciol):

what curve represents the plot of x(t) if the object is being accelerated with constant acceleration?

OpenStudy (fifciol):

or can you tell me what is the general form of equation of x(t)?

OpenStudy (anonymous):

ALL I know as of now is the behaviour of the car. Accelerating from rest means that the velocity is increasing, and the change in position is also positive, in the first quadrant. When the car stops, v =0, and the position time curve will be turning?But then , before he stops, he deccelerates, meaning that velocity was decreasing, and the p/t graph.

OpenStudy (anonymous):

but I am not sure about the equation @Fifciol

OpenStudy (vincent-lyon.fr):

I do not understand this part of the description: "and then returns to its original position slowing to a stop" isn't there anything missing here?

OpenStudy (anonymous):

@Vincent-Lyon.Fr - this is the question , as is.

OpenStudy (vincent-lyon.fr):

@Fifciol Are you also puzzled by this last bit?

OpenStudy (vincent-lyon.fr):

@MERTICH Have you tried drawing anything yet?

OpenStudy (fifciol):

I want you to figure out the shape of the curve. What I meant was what is the shape of the curve x(t) when object moves with constant acceleration. \[x(t)=x_0+v_0t+\frac{ 1 }{ 2 } at^2\] @Vincent-Lyon.Fr I think I know what he mean

OpenStudy (fifciol):

It doesn't have to accelarate with constant a, but this assumption may help you a bit to figure how to draw the plot

OpenStudy (anonymous):

oh, @Vincent-Lyon.Fr , you are right. A massive typo. its original position "BY SPEEDING UP TO A STOP. @Fifciol Yes, I know this equation very well. I just couldnt figure out where it will fit in the matter.

OpenStudy (anonymous):

.....its original position, by speeding up and then slowing to a stop, @Vincent-Lyon.Fr

OpenStudy (fifciol):

Can you draw the first part when car accelerates up to deccelaration starting point?

OpenStudy (vincent-lyon.fr):

"by speeding up and then slowing to a stop" Now this is clear. The was a missing step. So, @MERTICH, what types of graphs do you know, and what kind of motion do they represent? Please use the 'Draw' button to show us.

OpenStudy (vincent-lyon.fr):

Easier question: What would the graph look like if it was moving from zero to its return point with constant velocity?

OpenStudy (anonymous):

i know the velocity time graph, the acceleration -time graph, and the position time graph

OpenStudy (anonymous):

but getting it fro the acc/time graph is a challenge.

OpenStudy (vincent-lyon.fr):

This is not my question: show us what the graph is for constant velocity!

OpenStudy (anonymous):

|dw:1388856950923:dw|

OpenStudy (anonymous):

acc -increase in velocity,decelaration means decrease in velocity, stops at v=0,

OpenStudy (vincent-lyon.fr):

What I wanted you to draw is this: |dw:1388857305758:dw| That would be the graph if the first part and the second part of the motion were done at constant velocity. Now, you should combine your graph and mine to get to the real one.

OpenStudy (anonymous):

|dw:1388857530073:dw|

OpenStudy (anonymous):

it goes back to the real original position

OpenStudy (vincent-lyon.fr):

|dw:1388857604281:dw| No because, this part would mean that the body overshoots its starting position, which is not stated in the question.

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