The velocity of an airplane flying into a headwind is given by v(t) = 30(25-t²) mi/hr for 0 ≤ t ≤ 44 hr. Assume that s(0) = 0.a. Determine and graph the position function, for 0 ≤ t ≤ 44. b. How far does the airplane travel in the first 2 hr? c. How far has the airplane traveled at the instant its velocity reaches 400 mi/hr?

A. 30{25t-t³/3}miles

B. S(2) = 700miles

C. 1900√35/3miles

Explanation:

A. V(t) = 30(25-t²)

S(o) = 0

Velocity as a function of position

V(t) = ds(t)/dt where S(t) is the position function of the plane

Therefore, integral{ds(t)} = integral{V(t)dt = integral{30(25-t²)

= s(t)-s(o) = 30{25t-t³/3}(t,0)

= 30{25t-t³/3}miles

B. Distance of plane in 2 hours

S(2) = 30{25t-t³/3}

S(2) = 30{25(2)-2³/8}

S(2) = 30{50-8/3}

S(2) = 1500 - 2400/3

S(2) = 700miles.

C. V(t) = 30(25-t³) = 400

t² = 25 - 40/3

t = √35/3

S(√35/3) = 30{25-1/3(35/3)}√35/3

= {750-10×35/3}√35/3

= {750-350/3}√35/3

= 1900√35/3miles

Find attached below the position on the graph

it about 600 mph from what i calculated

neither accurate nor precise.