Answer to Question #200697 in Calculus for raf

Question

a ball is thrown down from the top of a 210-ft building with an initial velocity of -24 ft per second. the position function is s(t)= -16t^2+v_ot+s_o.

what is the velocity of the ball after 1 second.

find the rate of charge of y with respect to x on the internal [-2,2], where y=3x^2-2x

Answer 1

Given height of building is 210 feet.

initial velocity is -24 feet/second.

Given position function is s(t) = -16t²+v₀t+s₀

Differentiating both side with respect to x

ds(t)/dt = -32t + v₀

V(t) = -32t + v₀

Now velocity of the ball after 1 second is

V(1) = -32(1)+(-24)

V(1) = -56 feet/second.

Given y = 3x²-2x

Given interval is [-2,2]

Now rate of change of y with respect to x is

Differentiating y with respect to x

dy/dx = 6x-2

dy/dx at (-2) = 6(-2)-2 = -14

dy/dx at (2) = 6(2)-2 = 10

Thus, Rate of change of y with respect to x is [-14,10]