The fastest land animal is the cheetah, which runs at a speed of up to 101 km/h.
The second-fastest is the antelope, which runs at a speed of up to 88 km/h.
a) Suppose that a cheetah begin to chase an antelope. If the antelope has a head start of 50m, how long does it take the cheetah to catch the antelope: How far will the cheetah have travelled by this time?
b) The cheetah can maintain its top speed tor only about 20s (and then has to rest), whereas the antelope can continue at top speed for a considerably longer time. What is the maximum head start the cheetah can allow the antelope?
The speed of the cheeta $v_c = 101$ km/h
The speed of the antelope $v_a = 88$ km/h
The distance where they catch up is $s$
The time is $t$
The cheeta reaches this place after $s = t . v_c$
The antelope reaches this place after $s = 50 + t.v_a$
We have two equations that are equal to each other, so we can write:
$$t.v_c = 50 + t.v_a$$
$$\implies 50 = t.v_c - t . v_a$$
$$\implies 50 = t.(v_c - v_a)$$
$$\implies \frac{50}{v_c-v_a}= t$$
vc = 101 / 3.6 # m/s
va = 88 / 3.6 # m/s
t = 50 / (vc - va)
s = t * vc
s2 = 50 + t * va
print(f'Speed cheeta: {vc:.1f} m/s - Speed antelope: {va:.1f} m/s')
print(f'a) The cheeta catch up after {t:.2f} s at a distance of {s:.0f} m')
The cheeta can run $t_{max} = 20$ s at full speed and thus run a distance of $s_{max} = t_{max}.v_c$
On the same time, the antelope runs $s_{a} = t_{max}.va$
So if the Antelope has a a greater head start than $s_h = s_{max} - s_a$ the cheeta cannot cach-up any more.
t_max = 20
s_max = 20 * vc
s_a = 20 * va
s_h = s_max - s_a
print(f'The maximal distance the cheeta can run is {s_max:.0f} m')
print(f'in this time the antelope runs {s_a:.0f} m')
print(f'So if the antelope has a head start of more than {s_h:.0f} m, the cheeta has to order pizza.')