Five Steps

According to Physics legend Mr. Brian Fudacz, Physics is not math class and should be treated with high intellectual esteem. I don't really give a fuck but if we don't do these five steps we get points docked off on tests.

Steps

  1. Draw a picture of the scenario
  2. Identify all variables needed for the Kinematic Equations
  3. Choose and write down the unmodified equation variation
  4. Plug and chug for a solution
  5. Verify that the solution is valid and sensible for the scenario

Example

A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates with a constant upward acceleration of 2.00 m/s^2 until its engines stop at an altitude of 150 m.
  1. What is the maximum height reached by the rocket?
  2. How long after lift-off does the rocket reach its maximum height?
  3. How long is the rocket in the air?

Step 1

FIXME Draw shitty pictures in GIMP

Part 1

Part 1 represents when the rocket is being launched into the air.

Part 2

Part 2 represents when the rocket is slowing to its top of decent.

Part 3

Part 3 represents the rocket falling back to earth.

Step 2

Variables

Part 1
  • $V_i = 50.0 \text{ m/s}$
  • $a = 2.00 \text{ m/s^2}$
  • $D = 150 \text{ m}$
  • $V_f = \text{ ?}$
  • $t = \text{ ?}$
Part 2
  • $V_i = V_f$1)
  • $V_f = 0$
  • $a = -9.8 \text{ m/s^2}$2)
  • $D = \text{ ?}$
  • $t = \text{ ?}$
Part 3
  • $D = 150 + D$3)
  • $a = -9.8 \text{ m/s^2}$
  • $V_i = 0$
  • $t = \text{ ?}$

Answers

  1. $= D_1 + D_2$
  2. $= t_2$
  3. $= t_1 + t_2$

Step 3

Equations

  1. $V_f^2 = V_i^2 + 2aD$
  2. $V_f = V_i + at$
  3. $V_f^2 = V_i^2 + 2aD$ (Part 2)
  4. $V_f = V_i + at$ (Part 2)
  5. $D = V_it + \frac{1}{2}at^2$ (Part 3)

Step 4

Equation 1

  1. $V_f^2 = (50.0)^2 + 2(2.00)(150)$
  2. $V_f^2 = 2500 + 600$
  3. $V_f^2 = 3100$
  4. $\sqrt{V_f^2} = \pm\sqrt{3100}$
  5. $\boxed{V_f = 55.6 \text{ m/s}}$4)

Equation 2

  1. $55.6 = 50.0 + 2.00t$
  2. $5.6 = 2t$
  3. $\boxed{t = 2.8 \text{ secs}}$

Equation 3

  1. $0 = (55.6)^2 + 2(-9.8)D$
  2. $0 = 3091.36 + -19.6D$
  3. $-3091.36 = -19.6D$
  4. $\boxed{D = 157.7 \text{ m}}$

Equation 4

  1. $0 = 55.6 + -9.8t$
  2. $-55.6 = -9.8t$
  3. $\boxed{t = 5.7 \text{ secs}}$

Equation 5

  1. $307.7 = 0t + \frac{1}{2}(9.8)t^2$5)
  2. $307.7 = 4.9t$
  3. $\boxed{t = 7.9 \text { secs}}$

Step 5

  1. $\boxed{307.7 \text{ m}} = D$6) $\checkmark$
  2. $\boxed{8.5 \text{ secs}} = t + t$7) $\checkmark$
  3. $\boxed{16.4 \text{ secs}} = 8.5 + t$8) $\checkmark$
1)
part 1's answer
2)
gravity
3)
part 2's answer
4)
negative is discarded as it doesn't make sense in context
5)
gravity isn't negative here; in the context of the problem, negative time wouldn't make sense
6)
part 3's D value
7)
sum of the times of part 1 and 2
8)
part 3's answer

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