An athlete use his last 0.9s to jump from 40m to another position of 44m of a race.Find the linear acceleration of the athlete if it is assumed to be constant

To find the linear acceleration of the athlete, we need to use the equation of motion:

s = ut + (1/2)at^2

where
s is the displacement (change in position),
u is the initial velocity,
a is the acceleration,
and t is the time.

In this case, we know the initial position (s) is 40m, the final position is 44m, and the time (t) is 0.9s. The initial velocity (u) is not given, but we can assume it is zero since the athlete starts from rest when jumping.

Rearranging the equation, we have:

s = (1/2)at^2

Substituting the given values, we get:

44 m - 40 m = (1/2) a (0.9 s)^2

4 m = (1/2) a (0.81 s^2)

Now, let's solve for the linear acceleration (a):

a = (4 m) / [(1/2) (0.81 s^2)]

a = (4 m) / (0.405 s^2)

a ≈ 9.88 m/s^2

Therefore, the linear acceleration of the athlete, assuming it is constant, is approximately 9.88 m/s^2.