#### Answer

The gazelle is able to escape from the cheetah with 7.5 meters to spare.

#### Work Step by Step

We can find the gazelle's constant speed.
$v = a~t = (4.6~m/s^2)(5.0~s)$
$v = 23~m/s$
Note that the cheetah gains on the gazelle the whole time they are both running. We can find the distance the cheetah can run before it must stop.
$d = v~t = (30~m/s)(15~s) = 450~m$
To escape, the gazelle has 15 seconds to travel $450~m-170~m$ which is 280 meters. We can find the distance $x_1$ the gazelle travels during the 5.0 second acceleration period.
$x_1 = \frac{1}{2}at^2$
$x_1 = \frac{1}{2}(4.6~m/s^2)(5.0~s)^2$
$x_1 = 57.5~m$
We can find the distance $x_2$ the gazelle could run in the next 10 seconds.
$x_2 = v~t = (23~m/s)(10~s)$
$x_2 = 230~m$
The total distance the gazelle can travel in 15 seconds is 230 m + 57.5 m which is 287.5 meters. Since the gazelle only needed to run 280 meters to escape, the gazelle is able to escape from the cheetah with 7.5 meters to spare.