A cheetah is pursuing an impala. The impala is running in a straight line constant speed of 16 m/s. The cheetah is 10 m behind the impala running...

The initial speeds of the impala and the cheetah are 
16m/s and 20 m/s.

The cheetah has a decileration of
1m/s^2.  So the distance travelled by cheetah in t seconds is given by s(t) =
ut+(1/2)at^2, where u is the initial speed , a is the acceleration (or decileration ) of
the moving object. Applying this to the cheetah, u = 20m/s, a = -1m/s^2. So s(t) =
20t+(1/2)(-1)t^2. Or s(t) = 20t^2 -(1/2)t^2.

After t
seconds the distance covered bt impala at a constant speed of 16m/s is
16t.

 The initiial gap between impala and cheetah (cheetah
is behind) is 10m

Therefore after t seconds the  behind gap
between the cheetah and the impala = 10+16t - [20t-(1/2)t^2] = 10 -4t+(1/2)t^2. So the
cheetah is behind the impala by 0.5t^2 -4t+10 after t seconds.