If a distance-time graph (starting at (0,10) and ending at (10,0) has a negative linear trend, then what would the velocity-time and acceleration-time graphs look like?

1 Answer
Zack M.
Oct 24, 2015

The graphs for the functions, x=-t+10, v=-1, and a=0.

Explanation:

This is a derivatives problem, where v=(dx)/(dt) and a=(d^2x)/(dt^2). Since we are given that x(t) is negative linear, we can infer that the velocity function, v(t) will be a negative constant, and therefore the object is not accelerating, so a(t)=0.

If we want to graph these curves, we need to start by finding x(t), which we can identify by finding the slope and y-intercept and plugging them into the slope intercept form of a linear function, x(t)=mt+b. To find the slope, use the slope formula;

m=(y_2-y_1)/(x_2-x_1) = (0-10)/(10-0) = -1

We are conveniently given the y-intercept, (0,10), so;

x(t)=-t+10

Velocity can be found by taking the derivative, which in the case of a linear function is just the slope, m.

v(t)=-1

Find the velocity derivative to get acceleration.

a(t)=0

So our graphs look like;

enter image source here

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