Constant acceleration equations

These equations relate displacement, velocity, acceleration, and time, and apply under the following conditions:

the acceleration is constant
the motion is measured from t = 0
the fact that several variables are vectors is accounted for with appropriate plus and minus signs

v = v_o + at

x - x_o = v_o t + ½ a t²

x - x_o = ½ (v + v_o) t

v² = v_o² + 2 a (x - x_o)

The equations are derived from the definitions of acceleration, velocity, and displacement.

Example - Applying the equations

You're driving your red car. A black Porsche is in the right lane, stopped at the red light. You're traveling at 40 km/hr in the left lane, and the light turns green when you're 15 meters from the stop line. You continue at a constant speed of 40 km/hr and pass the Porsche, which accelerates from rest at a constant rate of 3 m/s², starting at the moment the light turns green.

(a) How far from the stop line do you pass the Porsche?

(b) When does the Porsche pass you?

(c) If a Boston police officer happens to get you and the Porsche on the radar gun at the instant the Porsche passes you, will either of you be pulled over for speeding? Assume the speed limit is 50 km/hr.

Lay out the problem

Define an origin - the stop line is a good choice.
Choose a positive direction - here, the direction you're traveling.
Use a consistent set of units - let's use MKS, with speeds in m/s.
Organize the data - write down everything you know.
Draw a diagram

You Porsche

x_o -15 m 0

v_o 11.11 m/s 0

a 0 3 m/s²

Part (a)

(a) How far from the stop line do you pass the Porsche?
At the instant you pass the Porsche, the x values (yours and the Porsche's) have to be equal. Using the general equation giving x as a function of time, write down two equations, one giving your position as a function of time and one giving the Porsche's position. The general equation is:
x = x_o + v_o t + ½ a t²
For you : x₁ = -15 + 11.11 t + 0
For the Porsche : x₂ = 0 + 0 + ½ (3) t² = 1.5 t²
Now what?

Set the equations equal, and solve for the time(s) when the two cars are at the same position. Then plug the time back into the x equation(s) to find the position.
x₁ = x₂
-15 + 11.11 t = 1.5 t²
Bringing everything to one side gives:
1.5 t² - 11.11 t + 15 = 0
Solve using the quadratic formula:

where a = 1.5, b = -11.11, and c = 15
This gives two values for t, t = 1.776 s and t = 5.631 s. What do these two values mean?
To get the answer to question (a), plug t = 1.776 into either of your x expressions. Check your answer: both equations should give the same value.
For you, at t = 1.776 s, x₁ = 4.73 m.
For the Porsche, at t = 1.776 s, x₂ = 4.73 m.
Part (b)

(b) When does the Porsche pass you?
We've already calculated this answer - it is the second solution to the quadratic we solved in part (a), t = 5.631 s. Round this off to t = 5.6 s.
Part (c)

(c) If a Boston police officer happens to get you and the Porsche on the radar gun at the instant the Porsche passes you, will either of you be pulled over for speeding? Assume the speed limit is 50 km/hr.
You're traveling at a constant speed of 40 km/hr, which is under the speed limit.
To figure out how fast the Porsche is going at t = 5.631 seconds, use:
v = v_o + a t = 0 + (3) (5.631) = 16.893 m/s.
Converting this to km/hr gives a speed of 60.8 km/hr, so the driver of the Porsche is in danger of getting a speeding ticket.