S(t)=40t^3-3t^5

Question

1. The position of an object is given by s(t)=40t^ 3- -3t^ 5 ,; 0 <= t <= 4 where s is in meters and t is in seconds. The graph of s is given below. Let a be the t-coordinate of the point on the graph of s where the object reaches its maximum velocity. Mark the approximate location of the point (a, s(a))

2.use calculus to determine the max velocity of the object. Explain why in 2-3 sentences

1. what is the position of the object at the time when the max velocity is reached? What is the sign of the acceleration before and after this time?

Answer 1

1. s'(t) = 120t² - 15t⁴; s''(t) = 240t - 60t³ = 60t(4 - t²) = 0; t = 0 or t = 2 s

Velocity v(t) = s'(t) = 15t²(8 - t²); v(0) = 0; v(2) = 240 m/s; v(4) = - 1920 m/s

2.Maximum velocity at t = 4 s and max v = 1920 m/s in negative direction

a = 4 s; (4, - 1920)