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Wize High School Grade 12 Physics Textbook > Kinematics

One-Dimensional Kinematic Equations

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One-Dimensional Kinematics


Velocity, displacement, and acceleration can be related using the kinematic equations of motion. These equations are different depending on type of the motion in the problem. Here, we introduce kinematic equations of motions with constant velocity and constant accelerations in one dimension.


Motion with a Constant Velocity

  • For constant velocity motions (a=0a=0):
x−x0=vt\boxed{x-x_0=vt}

Motion with a Constant Acceleration

  • For constant acceleration motions, there are three useful formulas:
x−x0=v0t+12at2\boxed{x-x_0=v_0t+\frac{1}{2}at^2}

v=v0+at\boxed{v=v_0+at}

vf2=v02+2a(x−x0)\boxed{v_f^2=v_0^2+2a(x-x_0)}


  • When solving a problem, we can use any of the above equations depending on the information given and asked for in the problem.

Wize Tip
  • Kinematic equation can be used when acceleration is constant i.e a does not change with particle motion.
  • In order to find the right equation to use, look for the missing variable.


Watch Out!
It is crucial to pick an appropriate sign for each of the variables in above equations since they are all vectors!



Steps to Take to Solve Any Kinematics Problem:

  1. Draw and label a diagram.
  2. Define the coordinate system (doesn't matter what you choose, as long as you are consistent).
  3. Identify the known values with units.
  4. Pick the right equation. Notice how if you know any 3 of Δx⃗,v⃗o,v⃗f,a⃗,t\Delta \vec{x}, \vec{v}_o, \vec{v}_f, \vec{a}, t , you can entirely solve the question!
  5. Solve for the quantity you want. Write down all your steps!
  6. Think about the final answer you got. Do the signs and value make physical sense?


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Example: Braking Distance


A car can decelerate at 4.0 m/s2m/s^2. If somebody is driving this car at 110 km/h, what will the stopping distance be for the driver?

1. Draw and label the diagram.



2. Define the coordinates! Since this is just a 1D problem, let's follow convention and say right is "+x" and left is "-x"

3. List the unknowns (with units). Pay attention to the signs of the values; they depend on our coordinate system we defined.
v⃗0=110km/h\vec{v}_0 = 110km/h
v⃗f=0m/s\vec{v}_f = 0 m/s
a⃗=−4.0m/s2\vec{a} = -4.0 m/s^2

We can already see that all the units are good, except for the km/h. We need to convert that into m/s.
v⃗0=110km/h×1000m1km×1h3600s=30.6m/s\vec{v}_0 = 110km/h \times \frac{1000m}{1km} \times \frac{1h}{3600s} = 30.6m/s
4. Choose the right equations. We are ultimately trying to solve for Δx⃗\Delta \vec{x} . Since we have three known quantities, we should be able to solve for everything in this question.

5. There is an equation to solve for the displacement. We can now solve for the value.
vf2=vo2+2a⃗Δx⃗Δx⃗=vf2−vo22aΔx⃗=(0)2−(30.6m/s)22(−4.0m/s2)Δx⃗=117mv_f^2=v_o^2+2\vec{a} \Delta \vec{x} \newline \Delta\vec{x}=\frac{v_f^2-v_o^2}{2a} \newline \Delta\vec{x}=\frac{(0)^2-(30.6 m/s)^2}{2(-4.0 m/s^2)} \newline \Delta\vec{x}=117 m

6. The displacement is in the positive direction, which means that we are displaced 117m to the right. That makes sense since we are travelling in that direction before we started braking. If we want the distance, it's just the magnitude of this value. Since we started with 2 s.f. in all our numbers, we can round off our answer to 120m.
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Example: Stopping Distance of a Driving Car


A car can decelerate at 4.0 m/s2m/s^2. If somebody is driving this car at 110 km/h and human reaction time is 0.45 seconds, what will the stopping distance be for the driver?


Solution:

First, let's convert the speed: 110 km/h = 30.6 m/s.

There are two parts to the motion: first, we continue for 0.45 seconds at the current speed. Then, we continue to move forward while decelerating.

First part: constant speed, time is known, we want distance.
x=vt=(30.6m/s)(0.45s)=13.8mx=vt=(30.6 m/s)(0.45 s)=13.8 m
Second part: the acceleration is in the opposite direction of the velocity, so it needs to be negative. We know the final speed, initial speed, and acceleration, but not time. We want distance.

vf2=vo2+2axx=vf2−vo22ax=(0)2−(30.6m/s)22(−4.0m/s2)x=117mv_f^2=v_o^2+2ax \newline x=\frac{v_f^2-v_o^2}{2a} \newline x=\frac{(0)^2-(30.6 m/s)^2}{2(-4.0 m/s^2)} \newline x=117 m
We add these two distance together now: xtotal=13.8m+117m=131mx_{total}=13.8m+117m=131m


Practice: Vertical Frog Jump

A frog jumps by pushing off the ground, and accelerates very quickly while its feet are still on the ground. Its center of mass lifts 3.0 cm while it is pushing off. The frog reaches a height of 40 cm off the ground before falling again. What acceleration did it have while launching itself off the ground?
A vehicle travelling 130 km/h passes a police car at rest. As soon as the vehicle passes the police car, the police car turns its lights on and begins to accelerate to its maximum speed of 140 km/h.

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If the police car accelerates at a constant acceleration for 12 seconds, how far will it have traveled before reaching its maximum speed?