Set 2 Problem number 6

If velocity increases by 6 meters per second per second, by how many meters per second does it increase in 7 seconds?

If velocity is initially 3 meters per second, then what is the velocity after 7 seconds?

What will be the average velocity and the distance traveled during the 7 seconds?

Solution

If every second velocity increases by 6 meters per second, then in 7 seconds it will increase by 42 meters per second.

Starting at 3 meters per second, this increase will result in a velocity of 45 meters per second.

The average velocity will thus be ( 3 + 45) / 2 meters per second = 24 m/s.

At this average velocity, the distance in 7 seconds will be ( 24 m/s)( 7 ) meters = 168 meters.

Generalized Solution

If the rate at which velocity changes is denoted a , then in time interval `dt the velocity changes by a `dt.

If the velocity starts at v0 and increases by a `dt, the velocity attained will be vf = v0 + a `dt.

Since the rate at which velocity changes is constant, the average velocity will be the average of the initial and final velocities:

vAve = ( v0 + vf ) / 2, where vf = v0 + a `dt (as above).

The displacement will therefore be `ds = vAve * `dt, using the vAve just obtained.

This sequence of calculations will get us to `ds.

We could obtain a single expression for `ds:

Substituting vf = v0 + a `dt into vAve = (v0 + vf) / 2 we get vAve = (v0 + [v0 + a `dt] ) / 2.
Simplifying this expression we have vAve = (2 v0 + a `dt) / 2 = v0 + a `dt / 2.
Using this expression in `ds = vAve * `dt we obtain `ds = (v0 + a `dt / 2) * `dt = v0 `dt + a `dt^2 / 2.

Explanation in terms of Figure(s), Extension

The 'blue' relationship in the first figure below shows what we know, that the displacement can be obtained from average velocity and time interval:

`ds = vAve * `dt.

The information given does not include average velocity.

However average velocity can be obtained for a uniform acceleration situation if initial and final velocities are known, as they are here.

The 'black' relationship shows that average velocity is obtained from v0 and vf.

The relationships among various variables, including the 'payoff' relationship `ds = (v0 + vf) / 2 * `dt, are shown in red.

We have reasoned out uniform accelerations situations using seven quantities: `ds, `dt, vAve, `dv, v0, vf and a.

We need to use all seven of these parameters to clearly think through any uniform motion problem.
However, there are some situations we cannot easily reason through (e.g., the situation in which we know a, `ds and v0).
For these situations we need to formulate a set of equations relating the various parameters.

Some of these parameters are directly related to others; e.g., `dv = vf - v0.
We can in fact formulate uniform acceleration in terms of just five independent parameters.
There are several possible choices for these parameters.
In what follows we will formulate uniformly accelerated motion in terms of the parameters `ds, `dt, v0, vf and a.

We characterize uniform acceleration by the five parameters `ds, `dt, v0, vf and a.

The present sequence of problems and associated figures has demonstrated two fundamental relationships among `ds, `dt, v0, vf and a:

`ds = (vf + v0) / 2 * `dt

a = (vf - v0) / `dt.

These two equations will be referred to as the two most basic equations of uniformly accelerated motion.

The five parameters are all we need to completely characterize any uniform acceleration situation.

We can solve any uniform-acceleration problem if we know only need three of these quantites.

Given any three of the quantities `ds, `dt, v0, vf and a we can find the values of the other two using the two equations listed above.
However, some situations would require solution of systems of linear equations.

Two more equations,

vf^2 = v0^2 + 6 `ds

and

`ds = v0 `dt + .5 a `dt^2,

can be derived from the first two.

The first equation is derived by eliminating `dt from the two most basic equations; the second by eliminating `ds from the two most basic equations.
The second figure below outlines the derivation of these equations.

The remaining two figures show how we might proceed in two situations:

In the first situation we assume that we know v0, vf and `dt.
In the second we assume that we know v0, a and `ds.

If we know v0, vf and `dt, we can easily reason out our solution in one of two ways:

Starting with v0 and vf, we can find either vAve (the 'black' path upward from vf and v0) or `dv (the 'blue' path downward from vf and v0).

If we have first found vAve, we can put it together with `dt to find `ds (the 'red' path).
If we have first found `dv, we put it together with `dt to get a.

Alternatively, we could have solved the problem using the equations:

Plugging our known values into equation E2 we immediately obtain `ds.
Having obtained `ds, we could solve either equation E1 or E3 for `dt.

(To avoid the quadratic equation E3 gives us for `dt we would almost certainly use the much simpler equation E1 to obtain `dt).

If we have a situation in which all we know is v0, a and `ds, there is no way to directly reason out vAve or `dv, or anything else.

We simply have to either set up a system of simultaneous equations using E1 and E2, or we use E3.
E3 is more convenient. That's why we derived it.
We solve equation E3 for vf, then plug in our information to get the value of vf.

This is indicated by the three 'blue' lines connecting v0, a and `ds to vf.

Having obtained vf, we then solve equation E2 for `dt.
Alternatively, at this point we can put vf together with v0 to get vAve. Then, from vAve and `ds we easily reason out `dt.

These relationships will be explored more fully in your class notes and your text. They are included here for your information.

Figure(s)

vf_and_dt_from_v0_a_and_ds.gif