Question

A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of a gazelle assumes an acceleration of 4.2 m/s2 for 6.5 s , after which the gazelle continues at a steady speed.What is the gazelle's top speed?A human would win a very short race with a gazelle. The best time for a 30 m sprint for a human runner is 3.6 s. How much time would the gazelle take for a 30 m race?
A gazelle would win a longer race. The best time for a 200 m sprint for a human runner is 19.3 s. How much time would the gazelle take for a 200 m race?

Answer 1

1. 27.3 m/s

The velocity of the gazelle at any time is given by:

$v=u+at$

where

u is the initial velocity

a is the acceleration

t is the time

Here we have:

u = 0 (the gazelle starts from rest)

$a=4.2 m/s^2$

t = 6.5 s

Substituting the data, we find the gazelle's top speed:

$v=0+(4.2)(6.5)=27.3 m/s$

2. 3.8 s

The distance covered by the gazelle is

d = 30 m

We know that the gazelle accelerates during the first part of the motion and then it continues at constant speed. We need to find first if the gazelle completes the race during the first part of its motion (accelerated motion); to do this, we can calculate what would be the distance covered by the gazelle before reaching the top speed, after t = 6.5 s:

$d'=\frac{1}{2}at^2 = \frac{1}{2}(4.2)(6.5)^2=88.7 m$

Which is larger than 30 m: this means that the gazelle covers the 30 m during its accelerated motion. Therefore, we can use again the equation:

$d=\frac{1}{2}at^2$

And substituting d = 30 m, we find the time:

$t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(30)}{4.2}}=3.8 s$

3. 10.6 s

In this case, the  distance the gazelle must cover is 200 m.

We know that in the first 6.5 s, the gazelle covers a distance of 88.7 m.

In the second part of the motion, the gazelle continues at its top speed, which is:

v = 27.3 m/s

The gazelle still have to cover a distance of

$d' = 200-88.7 =111.3 m$

Therefore, the time taken to cover this distance is

$t'=\frac{d'}{v}=\frac{111.3}{27.3}=4.1 s$

So, the total time the gazelle needs to cover 200 m is

t = 6.5 + 4.1 = 10.6 s