# Net Displacement (physics)

• September 28th 2010, 05:19 PM
mode
Net Displacement (physics)
Attachment 19101
the velocity-versus-time graph for a basketball player traveling up and down the court in a straight-line path. Find the net displacement of the player for the 10 s shown on the graph.

I am not too sure where to even start. Or even how someone achieves a velocity of -2m/s. Am i supposed to find his displacement on the X axis? if so, am i supposed to use this equation: $X=X_{xo}+V_{xo}+\frac{1}{2}a_{x}t^2$
• September 28th 2010, 05:56 PM
skeeter
Quote:

Originally Posted by mode
Attachment 19101
the velocity-versus-time graph for a basketball player traveling up and down the court in a straight-line path. Find the net displacement of the player for the 10 s shown on the graph.

I am not too sure where to even start. Or even how someone achieves a velocity of -2m/s. Am i supposed to find his displacement on the X axis? if so, am i supposed to use this equation: $X=X_{xo}+V_{xo}+\frac{1}{2}a_{x}t^2$

http://www.mathhelpforum.com/math-he...t-f131g2_r.jpg

The equation $\Delta x = v_0 t + \frac{1}{2}at^2$ is only valid if acceleration is constant. The given velocity graph clearly shows that acceleration is different over each lettered interval.

FYI, on a velocity vs. time graph, displacement is the area between the graph and the t-axis. Area above the t-axis is positive displacement, area below is negative displacement. Algebraically sum the positive and negative displacements to get the total displacement.
• September 28th 2010, 06:01 PM
mode
so i just sum up the velocity at each t interval?
• September 29th 2010, 03:29 AM
skeeter
Quote:

Originally Posted by mode
so i just sum up the velocity at each t interval?

one more time ...

on a velocity vs. time graph, displacement is the area between the graph and the t-axis. Area above the t-axis is positive displacement, area below is negative displacement. Algebraically sum the positive and negative displacements to get the total displacement.