Answer:
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
