Question

A cheetah is walking at a speed of 1.15 m/s when it observes a gazelle 43.0 m directly ahead. If the cheetah accelerates at 9.25 m/s2, how long does it take the cheetah to reach the gazelle if the gazelle doesn't move?

Answer 1

Answer:

2.93 s

Explanation:

Given that,

Initial speed of the Cheetah is, $u = 1.15 m / s$

distance has to be covered by Cheetah, $S= 43 m$

Acceleration of the Cheetah is,  $a= 9.25 m/s^{2}$

From the second equation of motion,

$S = ut + ( 1/ 2) at ^ 2$

Therefore,

$43 = 1.15 t + 4.625 t ^ 2\\4.625 t ^ 2 + 1.15 t -43 = 0$            

Therefore,

$t =\frac{ ({ -1.15\pm\sqrt{[ 1.15^ 2 - 4(4.625)(-43)}] })}{2( 4.625)}$

$t=\frac{-1.15\pm28.24}{2\times 4.625}$

Now only consider positive value of t.

$t=\frac{-1.15+28.24}{2\times 4.625}\\=2.93s$

Therefore the time taken by Cheetah to reach the gazelle is 2.93 s.