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course Phy 201
This is a resubmit because I didn't pay attention to my notes and missed the formula.
Possible Combinations of VariablesThere are ten possible combinations of three of the the five variables v0, vf, a, t and s. 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 t 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, t, s, v 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 # 4, 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.
A car has an initial velocity of 5m/sec and has an acceleration rate of 3m/sec^2. What will be the final velocity in 10 sec. What was the displacement of the car?
vf=a * ‘dt +v0
vf=3m/sec^2 * 10sec +5m/sec=35m/sec
vAve= (vf+ v0)/2
vAve=(5m/sec+35m/sec)/2=20m/sec
‘ds=vAve * ‘dt
‘ds= 20m/sec* 10sec= 200m
I tried to use the equations and tried to substitute different ones into other, but could not seem to find out specifically how to solve for the unkown variables in situation 5. I spent a good hour and a half but with no success. The steps include trying as you said to try and eliminate ‘dt from the equation. So to do this, I used the avg velocity formula and the acceleration formula. Then I tried substituting one into the other. But the units still did not work out right. I was pretty discouraged at this point and decided to move on.
(I feel like a dumbass now)
Looking at the formula for final velocity, I find that we can solve for this using
vf^2 = v0^2 + 2 * a * `ds
Then we can find ‘dt and vAve as well.
A toy car starts from rest and accelerates at a rate of 5cm/sec while traveling a total distance of 150cm. What is its final velocity and how long did it take to travel this distance?
vf^2 = v0^2 + 2 * a * `ds
vf^2 = (0) + 2 * 5cm/sec^2 * 150cm
vf^2 = 1500cm^2/sec^2
vf=38.7cm/sec
vAve=(38.7cm/sec + 0cm/sec) / 2
vAve= 19.4sec
‘ds = vAve * ‘dt
‘dt = 150cm / 19.4cm/^2
‘dt = 7.7sec
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&#Very good responses. Let me know if you have questions. &#
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