# Position from velocity time graph [closed]

I have taken Calculus before but unfortunately no equation is given so I have to do it the old way.

An object is at x = 0 at t = 0 and moves along the x axis according to the velocity-time graph shown below

I am asked to find the final position x of the object at t = 18s.

Here is my logic, but apparently the answer is wrong.

I find the area of each of the sections of the graph. If the graph is below the x axis it's area will be negative. I found the answer to be 97m. I got this from -44+-11+28.5+76+47.5. But this is wrong.

## closed as off-topic by John Rennie, Kyle Kanos, Daniel Griscom, Norbert Schuch, user36790 Jan 17 '16 at 2:07

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• since you have no intial condition in this question, you can't solve the problem. Please specify the initial condition. – Bruce Lee Jan 16 '16 at 16:42
• @BruceLee Sorry, An object is at x = 0 at t = 0 and moves along the x axis according to the velocity-time graph shown below – frillybob Jan 16 '16 at 16:43
• Hi and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. – John Rennie Jan 16 '16 at 16:48
• PS the initial velocity is -12 not -11 – John Rennie Jan 16 '16 at 16:50
• Use $d = d_0 + v_0t + 1/2 a t^2$ where $d_0$ is the initial displacement, $v_0$ is the initial velocity, $a$ is the slope of the curve (acceleration) and $t$ is the time interval. Do the same over all four different parts of the diagram and add them up. Be careful with the signs. – Bruce Lee Jan 16 '16 at 16:52