Question

A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 20 m/s when it reaches ground level. What was its speed when its height was half that of its starting point

Answer 1

Approximately
`14\; {\rm m\cdot s^(-1)}`.

Step-by-step explanation:

The gravitational potential energy
`\text{GPE}` of this roller coaster is proportional to the height
`h` of the roller coaster.

The kinetic energy
`\text{KE}` of this roller coaster is proportional to the square of speed
`v`.

The question states that the track is frictionless. Thus, during the descent, the
`\text{GPE}` of this roller coaster is turned into
`\text{KE}` without any energy loss.

When the roller coaster was at
`(1/2)` of the the initial height, only
`1 - (1/2) = (1/2)` of the original
`\text{GPE}` was turned into
`\text{KE}`. The
`\text{KE}\!` of this roller coaster at that height would be
`1 - (1/2) = (1/2)\!` of the
`\text{KE}\!\!` when the roller coaster is at the ground level.

The
`\text{KE}` of the roller coaster is proportional to
`v^(2)` (the square of speed
`v`.) Thus, since the
`\text{KE}\!` at
`(1/2)` the initial height is
`1 - (1/2) = (1/2)\!` the
`\text{KE}\!\!` at the ground level, the
`v^(2)` at
`(1/2)\!` the initial height would also be
`1 - (1/2) = (1/2)\!` the
`v^(2)\!` at the ground level.

Since
`v = 20\; {\rm m\cdot s^(-1)}` at the ground level,
`v^(2) = (20\; {\rm m\cdot s^(-1)})^(2)` at the ground level. The
`v^(2)` at
`(1/2)` the initial height would then be:

`(1 - (1/2))* (20\; {\rm m\cdot s^(-1)})^(2)`.

Thus, the speed
`v` at
`(1/2)` the initial height would be:

`\begin{aligned}& \sqrt{(1 - (1/2))* (20\; {\rm m\cdot s^(-1)})^(2)} \\ =\; & \sqrt{(1)/(2) * (20\; {\rm m\cdot s^(-1)})^(2)} \\ =\; & √(200)\; {\rm m\cdot s^(-1)} \\ \approx\; & 14\; {\rm m\cdot s^(-1)}\end{aligned}`.