Why is the slope of a distance velocity squared graph straight and a distance velocity graph is not?

Answer 1

When acceleration is constant, one equation of kinematics is:

(final velocity)^2 = 2(acceleration)(displacement) + (initial velocity)^2.

When you are graphing this equation with displacement or position of the x-axis and (final velocity)^2 on the y-axis, the equation becomes:

y = 2(acceleration)x + (initial velocity)^2.

Since acceleration is constant, and there is only one initial velocity (so initial velocity is also constant), the equation becomes:

y = constant*x + constant.

This looks strangely like the equation of a line:

y = mx + b.

Therefore, the slope of a velocity squared - distance graph is constant, or there is a straight line.

Now, when you graph a velocity - distance graph, the y axis is only velocity, not velocity squared. So if:

v^2 = mx + b.

Then:

v = sqrt(mx + b). Or: y = sqrt(mx + b).

This equation is not a straight line. For example, pretend m = 1 and b = 0. So the equation simplifies to:

y = sqrt(x).

Now, make a table of values and graph:

x | y

1 | 1

4 | 2

9 | 3

etc.

When you plot these points, the result is clearly NOT a straight line.

Hope this helps!