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course Phy 231
2-5-11 about 8:00pm
Possible Combinations of VariablesThere are ten possible combinations of three of the the five variables v0, vf, a, Δt and Δs. These ten combinations are summarized in the table below:
1 v0 vf a
2 v0 vf dt
3 v0 vf ds
4 v0 a dt
5 v0 a ds *
6 v0 dt ds
7 vf a dt
8 vf a ds *
9 vf dt ds
10 a dt ds
If we know the three variables we can easily solve for the other two, using either direct reasoning or the equations of uniformly accelerated motion (the definitions of average velocity and acceleration, and the two equations derived from these by eliminating Δt and then eliminating vf).
Only two of these situations require equations for their solution; the rest can be solved by direct reasoning using the seven quantities v0, vf, a, Δt, Δs, Δv and vAve. These two situations, numbers 5 and 8 on the table, are indicated by the asterisks in the last column.
Direct Reasoning
We learn more physics by reasoning directly than by using equations. In direct reasoning we think about the meaning of each calculation and visualize each calculation.
When reasoning directly using v0, vf, `dv, vAve, `ds, `dt and a we use two known variables at a time to determine the value of an unknown variable, which then becomes known. Each step should be accompanied by visualization of the meaning of the calculation and by thinking of the meaning of the calculation. A 'flow diagram' is helpful here.
Using Equations
When using equations, we need to find the equation that contains the three known variables.
• We solve that equation for the remaining, unknown, variable in that equation.
• We obtain the value of the unknown variable by plugging in the values of the three known variables and simplifying.
• At this point we know the values of four of the five variables.
• Then any equation containing the fifth variable can be solved for this variable, and the values of the remaining variables plugged in to obtain the value of this final variable.
Problem
Do the following:
• Make up a problem for situation # 7, and solve it using direct reasoning.
• Accompany your solution with an explanation of the meaning of each step and with a flow diagram.
• Then solve the same problem using the equations of uniformly accelerated motion.
Make up a problem for situation # 5, and solve it using the equations of uniformly accelerated motion.
An object accelerates uniformly at .3cm/sec/sec over a clock interval of 15sec at which it has a final velocity of 45cm/sec. What is the v0, dv, ds, dt, and vAve of the object?
aAve= .3cm/sec/sec
vf= 45cm/sec
dt= 15sec
v0= .3cm/sec/sec * 15sec= 4.5cm/sec, subtract this from 45cm/sec and you get 40.5cm/sec
dv= 45cm/sec – 40.5cm/sec= 4.5cm/sec
vAve= (45cm/sec + 40.5cm/sec)/2= 42.75cm/sec
ds= 42.75cm/sec * 15sec= 641.25cm
Step 1) since we know what vf, dt and a is we can find v0 by multiplying dt by a. This product is then subtracted from vf which gives us v0.
Step 2) Having v0 and vf, we can find dv
Step3) Having v0 and vf, we can find vAve
Step4) Knowing vAve, we can find the ds
Using equations of uniformly accelerated motion we can solve for v0. The equation vf= v0 + a * dt can be used to find v0 by solving for it. This gives the equation v0= vf – a * dt. Now using (v0 + vf)/2 * dt we find ds.
An object starts out with a velocity of 12m/sec, it increases its velocity at a rate of 2m/sec/sec for 80m. What is the vf, dv, ds, dt, vAve?
vf^2= v0^2 + 2(a)(ds)
= (12m/sec)^2 + 2(2m/sec/sec)(80m)
= 144m^2/sec^2 + 320m^2/sec^2
= 464m^2/sec^2
take the square root of both sides and you get vf= 21.54m/sec
Then use ds= (v0 + vf)/2 *dt to find dt
80m= (12m/sec + 21.54m/sec)/2 * dt
80m= 16.77m/sec *dt
4.77sec= dt
vAve= 16.77m/sec taken from the above calculation . . . (12m/sec + 21.54m/sec)/2
dv= 21.54m/sec – 12m/sec= 9.54m/sec
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