Verified answer

Is the whole 3 + 6t inside the square root? If so then find ds/dt

s = (3 + 6t) ^ 1/2

ds/dt = 1/2 (3 + 6t) ^ -1/2 • 6

= 3 / √(3 + 6t)

which when t = 1 would be 3 / √9 = 3/3 = 1 m/s

a) a(t) = dv/dt v(t) = ds/dt v(t) = ds/dt = 4t^3 - 100t ft/sec a(t) = dv/dt = 12t - one hundred ft/sec^2 b) s(2) = 2^4 - 50 * 2^2 + 625 = sixteen - 2 hundred + 625 = 441 ft. v(2) = 4*2^3 - one hundred * 2 = 32 - 2 hundred = -168 ft/sec a(2) = 12 * 2 - one hundred = 24 - one hundred = -seventy six ft/sec^2 c) locate the factors the place v(t) = 0 0 = 4t^3 - 100t 0 = t^3 - 25t 0 = t(t+5)(t-5) t = 0 t+5 = 0 :: t = -5, no longer a answer considering this fee is below 0. t-5 = 0 :: t = 5 so which you're stopped at 0 sec. and 5 sec. d) whilst does a(t) = 0? 12t - one hundred = 0 12t = one hundred t = 25/3 We desperate that the acceleration at 2 seconds replaced into adverse already, so it decreases interior the era [0,25/3), and will boost at (25/3, infinity). 25/3 isn't blanketed in the two era with the aid of fact a = 0.