The equations of constant acceleration.

Step 1 of 6

Position from the area

With constant acceleration the velocity graph is a straight line from $v_0$⁠ up to $v = v_0 + at$⁠. The area under it, up to a time $t$⁠, is a rectangle with a triangle on top:

• The rectangle: height $v_0$⁠ and width $t$⁠, area $v_0 t$⁠.
• The triangle: base $t$⁠ and height $at$⁠, area $\tfrac12 \cdot t \cdot at = \tfrac12 at^2$$\tfrac12 \cdot t \cdot at =$$\tfrac12 at^2$⁠.

The area is the displacement, so the position is

$x = x_0 + v_0 t + \tfrac12 at^2$$x =$$x_0 + v_0 t + \tfrac12 at^2$⁠

where $x_0$⁠ is the starting position. In the scene $x_0 = 0$⁠. When the acceleration is negative the triangle is negative too, and it takes away from the displacement.

Step 2 of 6

When it stops

At the turning point the velocity is zero. So its time comes from the velocity equation, with $v = 0$⁠:

$0 = v_0 + at$⁠

That is a linear equation with one unknown: $t$⁠.

Step 3 of 6

Stopping on a mark

Together the two equations let you plan a run. If the cart has to stop at a given time, $v_0 + at = 0$⁠ ties the starting velocity to the acceleration. If it also has to stop at a given place, $x = v_0 t + \tfrac12 at^2$$x =$$v_0 t + \tfrac12 at^2$⁠ fixes both. Work it out first, and only then move Starting velocity and Acceleration.

Step 4 of 6

Without the time

Sometimes the time is unknown and does not matter, for example when you want to know how many metres a car needs to stop. Solve the velocity equation for $t$⁠ and put that into the position equation, and you are left with an equation without time:

$v^2 = v_0^2 + 2a\,\Delta x$$v^2 =$$v_0^2 + 2a\,\Delta x$⁠

When the object stops, $v = 0$⁠, and the stopping distance is $\Delta x = \dfrac{v_0^2}{2|a|}$$\Delta x =$$\dfrac{v_0^2}{2|a|}$⁠. For the run on screen, 0.6 m/s and −0.3 m/s², that is $\dfrac{0.36}{0.6} = 0.6$$\dfrac{0.36}{0.6} =$$0.6$⁠ metres: exactly the top of the arch.

Step 5 of 6

Twice the speed

Now the cart crosses the start line at 0.3 m/s, and the fan pushes back at −0.3 m/s². It stops after one second, 0.15 m from the start line. In a moment it will set off at twice the speed, with the same push. Predict how far it will get before it stops.

Step 6 of 6

Summary

For motion with constant acceleration $a$⁠, from position $x_0$⁠ with starting velocity $v_0$⁠:

$v = v_0 + at \qquad x = x_0 + v_0 t + \tfrac12 at^2 \qquad v^2 = v_0^2 + 2a\,\Delta x$$v =$$v_0 + at \qquad x =$$x_0 + v_0 t + \tfrac12 at^2 \qquad v^2 =$$v_0^2 + 2a\,\Delta x$⁠

• Choose the equation by what you know and what you want: without the time, the third one.
• At the turning point $v = 0$⁠, and the stopping distance is $\dfrac{v_0^2}{2|a|}$⁠: twice the speed, four times the distance.
• In the projectile motion Unit the same equations describe a ball’s vertical motion, with the acceleration of free fall, $g$⁠.