Question

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?

Answer 1

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;