# Kinematics/Linear Motion

A reader requests expansion of this page to include more material.You can help by adding new material ( learn how) or ask for assistance in the reading room. |

## Problems[edit]

- 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 10 s 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 esculator at the same rate he was walking before?

Prerequisite: Calculus

- 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 cocroach from the wall after 1 s?

## Solutions[edit]

1. 3 s. Use the kinematics equation y = y0 + v0t + 1/2 at^2. Let the rocks 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 yeilds the solution immediately. Another route to the solution is to solve the differential equation by seperating 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.