course Phy 201 ProblemDo the following: • Make up a problem for situation # 3, 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 begins to roll down a hill at an initial velocity of 1 cm/s. The ball rolls for a time interval of 4 seconds before stopping at a final velocity of 4 cm/s. How far does the ball travel during this time interval? V0 = 1 cm/s Vf = 4 cm/s `dt = 4 s Therefore, vAve = 4 cm/s + 1 cm/s = 5 cm/s / 2 = 2.5 cm/s With initial velocity and final velocity given we were able to obtain the average velocity. Now with vAve and `dt we can find `ds. `ds = 2.5 cm/s * 4 s = 10 cm The ball rolls 10 cm in 4 s. V0 vf `dt vAve `ds The flow diagram shows that when initial velocity and final velocity are given in the first row we can find the average velocity in the second row. With the time interval also given in the first row we are able to obtain the change in position from the multiplication of the time interval and the average velocity. V0 = 1 cm/s Vf = 4 cm/s `dt = 4 s vAve = 4 cm/s + 1 cm/s = 5 cm/s / 2 = 2.5 cm/s `ds = 2.5 cm/s * 4 s = 10 cm `dv = 4 cm/s – 1 cm/s = 3 cm/s A = 3 cm/s / 4 s = 0.75 cm/s/s Using the equations of uniform accelerated motion find the v0 and `dt for an object that accelerates through a distance of 75 cm ending at velocity 7 cm/s and accelerating at 0.84 cm/s/s. `ds = 75 cm Vf = 7 cm/s A = 0.84 cm/s/s To find initial velocity we have to use the equation Vf^2 = V0^2 + 2a *`ds and solve for v0. Therefore we get, -v0^2 = 2a * `ds – vf^2 = 2(0.84 cm/s^2)(75 cm) – (7 cm/s)^2 = 126 cm^2/s^2 – 49 cm^2/s^2 Then take the square root of both sides to get 8.77 cm/s. However, we must then divide both sides by -1 to get an initial velocity of -8.77 cm/s. This means that the initial velocity is 8.77 in the left direction. Now that we have the initial velocity we can find the remaining variables. `dv = 7 cm/s - -8.77 cm/s = 15.77 cm/s vAve = 7 cm/s + -8.77 cm/s /2 = -0.885 cm/s `dt = 75 cm/-0.885 cm/s = 84.75 s