Answer:
1
[tex] t_a = 13.49 \ s [/tex]
2
The distance is [tex] D = 55.2 \ m [/tex]
Explanation:
From the question we are told that
The maximum speed of the cheetah is [tex]v = 103 \ km/h = 28.61 \ m/s[/tex]
The maximum of gazelle is [tex]u = 76.5 \ km/h = 21.25 \ m/s[/tex]
The distance ahead is [tex]d = 99.3 \ m[/tex]
Let [tex]t_a[/tex] denote the time which the cheetah catches the gazelle
Gnerally the equation representing the distance the cheetah needs to move in order to catch the gazelle is
[tex] v* t_a = d + u* t_a [/tex]
=> [tex] 28.61 t_a = 99.3 + 21.25t_a [/tex]
=> [tex] 7.36 t_a = 99.3 [/tex]
=> [tex] t_a = 13.49 \ s [/tex]
Now at t = 7.5 s
[tex]7.5 v = D+ 7.5u [/tex]
=> [tex] 28.61 * 7.5 = D + 21.25* 7.5 [/tex]
=> [tex] 7.36 * 7.5 =D [/tex]
=> [tex] D = 55.2 \ m [/tex]
Hence the for the gazelle to escape the cheetah it must be 55.2 m
