The highest speed of a cheetah is 100 km/hr and the highest speed of a gazelle is 80 km/hr. If at t = 0 both animals are running at their respective highest speeds, with the gazelle a distance 100 m ahead of the cheetah, at what time does the cheetah catch up with the gazelle? Both the cheetah and the gazelle run in the same direction in a straight line.

The final position of an object moving at constant speed is given by the formula [tex]x=x_0+vt[/tex], where [tex]x_0[/tex] is its initial position, v its speed and t the time elapsed.

For the cheetah we have [tex]x_c=x_{0c}+v_ct[/tex], and for the gazelle [tex]x_g=x_{0g}+v_gt[/tex]. We want to know at which t their positions are equal, that is, [tex]x_c=x_g[/tex], which means,

[tex]x_{0c}+v_ct=x_{0g}+v_gt[/tex]

Where we can do:

[tex]v_ct-v_gt=x_{0g}-x_{0c}[/tex]

[tex](v_c-v_g)t=x_{0g}-x_{0c}[/tex]

[tex]t=\frac{x_{0g}-x_{0c}}{v_c-v_g}[/tex]

We then substitute the values we have (the initial position of the cheetah is 0m), writing the meters in km so distance units cancel out correctly:

[tex]t=\frac{0.1km-0km}{100km/hr-80km/hr}=0.005hr=18s[/tex]

On the last step we just multiply by 3600 because is the number of seconds in an hour.