course phy 201
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Quiz 3 Version 1Do the follo
• Make up a problem for situation # 10, 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.
A ball is traveling at a constant acceleration of 0.444cm/s/s for 9second and traveled a distance of 117cm. Find the initial and final velocity.
Since you have the time and the displacement you can use the equation change in displacement = the initial velocity multiplied by the change in time plus 0.5 time the acceleration times the change in time squared. From this equation you can find the initial velocity. You then have to solve for the change in velocity. You can do this by using the average acceleration equals the change in velocity divided by the change in time. So to obtain the change in velocity you would multiply the acceleration by the time and you would get the change in velocity. You can then find the final velocity from this by adding the change in velocity to the initial velocity.
‘ds ‘dt a * don’t know how to draw lines
‘ds = vo*’dt + 0.5 *a*’dt^2 a=’dv/’dt ‘dv = vf-vo
‘ds= vo * ‘dt + 0.5*a*’dt^2 vo = (0.5 *a*’dt^2 – ‘ds)/-‘dt
(0.5 * 0.444cm/s * 9s^2 – 117cm)/ -9s
vo = 11cm/s
aAve = ‘dv/’dt ‘dv = aAve * ‘dt
= 0.444cm/s * 9s
‘dv = 4cm/s
‘dv = vf-vo vf = ‘dv + vo
4cm/s + 11cm/s
vf = 15cm/s
number 8 A ball traveled for 117cm at an average acceleration at 0.444cm/s. The final velocity was 15cm/s. What is the initial velocity and what is duration of time?
vf^2 = v0^2 + 2 * a * `ds vo = sqt(vf^2 – 2a’ds)
= sqt(15cm/s^2 -2*0.44cm/s/s *117cm)
vo = 11cm/s
vAve = (vf+vo)/2 vAve = (vf+vo)/2
= (15cm/s + 11cm/s)/2
vAve = 13cm/s
vAve = ‘ds/’dt ‘dt = ‘ds/vAve
= 117 / 13cm/s
‘dt = 9s
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