course phy201
6/24 3
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 Dt 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, Dt, Ds, Dv 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 # 8, and solve it using the equations of uniformly accelerated motion.
Problem 1
If we know the initial velocity of an object is 6cm/s and that it travels for 30 s. accelerating at a rate of 2cm/s^2 than what would be the change in position and the final velocity of the object?
reasoning:
since we know the acceleration and time intervals we can multiply these two together and add this to the initial velocity to determine the final velocity. 30s * 2cm/s = 60cm/s + 6cm/s= 66cm/s
now that we know the initial and final velocities we can average them which comes to 36cm/s. then take the average velocity and multiply this by the time interval. So the change in position is 1080cm.
formulas:
to determine `dv: a= `dv/`dt so 2cm/s^2= `dv/30= 60cm/s
to determine vF: `dv=vf-v0 so 60cm/s= vf- 6cm/s so vf= 60cm/s+6cm/s= 66cm/s
to determine average velocity: vAve= (vf+v0)/2 so (66cm/s+6cm/s)/2 = 36 cm/s
to determine `ds: vAve= `ds/`dt so 36cm/s= `ds/30 so 36cm/s *30s= 1080 cm
Problem 2
If a car travels 25 miles at an acceleration rate of 20m/h^2 and it's final velocity is 60m/h what is the car's initial velocity and how long does it take for the car to travel this distance?
to determine v0: vf^2= v0^2 + 2a `ds
60m/h^2= v0^2+ 2 * 20m/h^2 * 25 m
3600m/h= v0^2+ 1000m/h^2
2600 m/h=v0^2
sqrt2600m/h= sqrtv0^2
51m/h=v0
to determine`dt: a= `dv/`dt
20m/h^2=(60m/h- 51m/h)/ `dt
20m/h^2( `dt)=(9m/h)
`dt= (9m/h)/(20m/h^2)
`dt= .45hr"
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