Question

A roller-coaster car rolls down a frictionless track, reaching speed v0 at the bottom. If you want the car to go twice as fast at the bottom, by what factor must you increase the height of the track? Explain.

Short answer

As the speed depends only on the vertical height and not the shape of the track, the height of the track is increased by two times.

Step-by-step solution

Content Introduction

As we know the track is frictionless, let us consider the car and Earth as an isolated system. So, firstly find the equation of the final velocity at the bottom, when the car starts from the rest at height h.

$$\begin{gathered} ∆E=0 \\ {E}_f=E_i \\ {K}_f+U_{g,f}=K_i+U_{g,i} \\ \frac{1}{2}m{v^2}_{f}mg{y}_f=\frac{1}{2}m{v_i}^2+mg{y}_i \end{gathered}$$

Note that car starts from rest, $v_i=0$, $y_f=0$because it is at ground and$y_i=h_1$

$h_i=\frac{{v^2}_f}{g}$

Content Explanation

To make car go twice as fast at the bottom, we will use above equation.

$$\begin{gathered} h_2=\frac{{v^2}_{f_2}}{g} \\ Note:v_{f_2}=2v_f \\ {h}_2=\frac{{\left(2v_f\right)}^2}{g} \\ {h}_2=\frac{4{v^2}_f}{g} \\ {h}_2=4h_1 \end{gathered}$$