Uniformly Accelerated Motion

Definition of IB symbols:

u initial velocity
v final velocity
a acceleration (constant)
t time taken
s distance traveled

2.1.7 Derive the equations for uniformly Accelerated Motion

Acceleration is defined as the rate of change of velocity with respect to time, in a given direction. The SI units of acceleration are ms-2. This would mean that if an object has an acceleration of 1 ms-2 it will increase its velocity (in a given direction) 1 ms-1 every second that it accelerates. If we write the definition for acceleration in mathematical terms:

\begin{align} a ={v-u\over t} \end{align}

Then solve of v:

\begin{equation} v=u+at \end{equation}

This gives us the final velocity of an object in terms of the acceleration, initial velocity and the time the object was accelerating. In graphical form the equation is a straight line, with u as the vertical intercept, a as the slope and t as the independent variable.


We can derive yet another equation if we look at the graph of a velocity vs. time for an object with a non-zero initial velocity.

We know that the area under the curve is equal to the distance travele

\begin{equation} Area = Red + Yellow \end{equation}
\begin{align} s = ut + {(v-u)\over 2} t \end{align}
\begin{align} s = ut + {1\over 2} vt - 1{\over 2} ut \end{align}
\begin{align} s = {u+v\over 2} t \end{align}

In we now combine the two equations that we have derived, we can create another equation:

\begin{align} s = {u + u + at\over 2}t \end{align}
\begin{align} s = ut + {1\over 2} at^2 \end{align}

Now if we go back to the definition of acceleration (1) and multiply by t and divide by a:

\begin{align} t = {v-u\over a} \end{align}

If we substitute this expression for time into the 3rd equation and solve for v:

\begin{align} s=u \left [{v-u\over a} \right ]+{1\over 2} a \left [{v-u\over a} \right ]^2 \end{align}
\begin{align} s = {uv -u^2\over a} + { v^2 +u^2 -2uv \over 2a} \end{align}
\begin{equation} 2as = 2uv -2u^2 + v^2 +u^2 -2uv \end{equation}
\begin{equation} 2as = v^2 -u^2 \end{equation}
\begin{equation} v^2 = u^2 + 2as \end{equation}

So what do we have? We have four equations that describe uniformly accelerated motion:

Equation (2) provides a way of calculating the final velocity in terms of the initial velocity, acceleration and the time the object was accelerated.

Equation (6) provides a way of calculating the distance (displacement) of an object in terms of the initial velocity, final velocity and time the object was in motion.

Equation (8) gives us the distance traveled without having to know the final velocity of the object. In exchange for knowing the final velocity we must know the acceleration of the object.

Equation (14) relates the initial velocity, final velocity and acceleration of the object without time! Sometimes this is very useful.

It is very important to note that these equations ONLY apply if the object in question is experiencing uniform acceleration, meaning that the acceleration is constant or can be approximated as constant.

2.1.8 Describe the vertical motion of an object in a uniform gravitational field

A uniform gravitational field simply means that the force of gravity does not change or does not change significantly. If you stay near the surface of the Earth (with in a kilometer or two) the gravitational field can be assumed to be constant (at least for the purposes of IB physics).

If the gravitational field is constant then an object in that field will experience a constant force and thus a constant acceleration. This means that the equations derived above are a valid description of the motion of an object in a uniform gravitational field (assuming no other forces).

If an object is held stationary in a uniform gravitational field it will fall. It will do so with uniform acceleration. Near the surface of the earth the acceleration is approximately 9.8 ms-2. This means that every second that the object falls its velocity will increase by 9.8 ms-1. So after one second the object has a velocity of 9.8 ms-1, after 5 seconds it will have a velocity of 47.5 ms-1, etc. Since the objects velocity is increasing every second this naturally means that the distance it is covering each second also increases…

What happens if an object is thrown up? The acceleration is still downward. If an object is thrown up with an initial velocity of 30 ms-1, after one second it will only be going 20 ms-1 up, after 2 seconds it will only be going 10 ms-1, after 3 seconds the object will have zero velocity! Even if the objects velocity is zero the acceleration is not zero.

Vertical motion in a gravitational field will be discussed at more length when we get to projectile motion.

2.1.9 Describe the effects of air resistance of a falling object

As an object falls it experiences a drag or frictional force due to air, we call this force air resistance. This drag force is always in the opposite direction of the motion. When the object is moving slowly the force of drag is proportional to the speed of the object. As the object’s speed increase the drag is proportional to the square of the speed, which means that the drag force becomes large very quickly. At some point the drag force will increase to the size of the force of gravity on the object. When the two forces are equal their will be no net force on the object and it will no longer accelerate, i.e. it will now travel at a constant velocity. This final maximum velocity is called the terminal velocity. The drag and thus the terminal velocity is determined by the shape of the object, its mass and the cross section of the object.

People have survived falling out of airplanes and hot air balloons, this doesn’t happen often but it does happen once in a great while. They many fall for a couple of minutes before hitting the ground, where they actually bounce. It happens that a falling human body will reach its terminal velocity (about 200-300 km/hr) fairly quickly, so while they may fall for a long period of time they hit the ground at a relatively low speed. Low compared to the speed they would hit with if the air did not reduce their acceleration.

Want to add to or make a comment on these notes? Do it below.

Add a New Comment
or Sign in as Wikidot user
(will not be published)
- +
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License