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Problems[edit | edit source]
- 1. A man fires a rock out of a slingshot directly upward. The rock has an initial velocity of 15 m/s. How long will it take for the rock to return to the level he fired it at?
- 5 It takes a man ten seconds to ride down an escalator. It takes the same man 15 seconds to walk back up the escalator against its motion. How long will it take the man to walk down the escalator at the same rate he was walking before?
- 24. Starting with the definitions of velocity and acceleration derive the kinematics equation for constant acceleration x = x0 + v0t + (1/2) at^2.
- 25. A cockroach starts 1 cm away from a wall. It starts running in a line directly away from the wall with a velocity of 1 cm/s, acceleration of 1 cm/s^2, jerk (d^3 x/dt^3) of 1 cm/s^3, d^4 x/dt^4 of 1 cm/s^4, and so on. How far is the cockroach from the wall after one second?
Solutions[edit | edit source]
1. 3 s. Use the kinematics equation y = y0 + v0t + 1/2 at^2. Let the rock's initial height y be 0. Plugging the values into the equation yields 0 = 15t - (1/2) 10t^2. Factoring and solving for t yields solutions of 0 s and 3 s. Hence it takes the rock 3 s to return to its original height.
24. The quickest solution is to write the Taylor series for x, this yields the solution immediately. Another route to the solution is to solve the differential equation by separating the variables.
25. e cm. The Taylor series for positions is x = x0 + v0 t + (1/2!) a0 t^2 + (1/3!) j0 t^3 + ... . Since x0, v0, a0, ... are all equal to one and t, t^2, t^3, ... are all equal to one, you have the following series: x = 1 + 1 + 1/2 + 1/6 + ... . This is a series that sums to e, thus the cockroach travels e cm after one second.