Question

A softball player chasing a fly ball runs in a line towards the center field fence. Suppose the player's velocity is v(x)=0.25x^2-1.10x+25 when the player is located at the point x in the field. What is the acceleration when the player is 2 meters from the fence?

Answer 1

understand that differentiating velocity would give you acceleration.
v'(x) = 0.5x - 1.10
when x =2,
v'(2) = 0.5(2) - 1.1 = 0.1 - 1.1 = -1 m/s^2

Answer 2

velocity v(x) = 0.25x^2-1.10x+25 when distance = x
the acceleration = v'(x) = 2*(0.25x) - 1.10
and if distance = 2 meters then
the acceleration will be = 2*[0.25*2] -1.10 = -0.1meters/second^2
the acceleration = -0.1 m/s^2