Question

The cheetah can maintain its maximum speed
for only 7.5 s.
What is the minimum distance the gazelle
must be ahead of the cheetah to have a chance
of escape? (After 7.5 s the speed of cheetah is
less than that of the gazelle.)
Answer in units of m.

Answer 1

A cheetah can run at a maximum speed 108 km/h and a gazelle can run at a maximum

speed of 75.9 km/h.

The velocity of the cheetah in meters per second is:

$V_{oc}=108\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}\\ V_{oc}=30\frac{m}{s}$

The velocity of the gazelle in meters per second.

$V_{og}=75.5\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}\\ V_{og}=21.1\frac{m}{s}$

Establishing a reference system where initial the cheetah.

The position of the cheetah is given by:

$X_{C}=X_{oc}+V_{oc}*t$

Where:

• Initial position $X_{0}=0m$
• Time $t=7.5s$
• velocity $V_{oc}=30\frac{m}{s}$

Expression for the position.$X_{C}=V_{oc}*t$

The position for the gazelle is given by:

$X_{g}=X_{og}+V_{og}*t$

Where:

• Initial position $X_{og}=??$
• Velocity of the gazelle $V_{og}=21.1\frac{m}{s}$
• Time $t=7.5s$

Expression.$X_{g}=X_{og}+V_{og}*t$

When the cheetah barely reaches the gazelle it is fulfilled that:

$X_{c}=X_{g}$

replacing expressions

$V_{oc}*t=X_{og}+V_{og}*t$

resolving for the initial position of the gazelle

$V_{oc}*t=X_{og}+V_{og}*t \\ V_{oc}*t=X_{og}+V_{og}*t \\ X_{og} = V_{oc}*t-V_{og}*t \\ X_{og} = 30\frac{m}{s}*7.5s-21.1 \frac{m}{s}*7.5s \text{ Numerically evaluating }\\X_{og}=66.75m$

the minimum distance the gazelle must be ahead of the cheetah to have a chance of escape:

$\boxed{X_{og}=66.75m}$