# Chapter 2 - Derivatives - 2.1 Derivatives and Rates of Change - 2.1 Exercises: 14

a)6.82m/s b)v = 10 - 3.72a c)t = 5.376s d)v = -10m/s

#### Work Step by Step

a) To find the velocity of the rock after one second, we need to figure out the equation for velocity. Velocity is the change in position with respect to time, so the equation for the rock's velocity is the derivative of the position function H. If $H = 10t - 1.86t^{2}$ Then $H' = V = 10 - 3.72t$ We can then plug in 1 for t to get 6.28m/s b) To find the velocity of the rock at t = a, we simply plug in a for t to get: v = 10 - 3.72(a) c) The rock will hit the surface when the position(H) equals 0. $0 = 10t - 1.86t^{2}$ $0 = t(10 - 1.86t)$ $0 = (10 - 1.86t)$ t = 5.376 d) To find the velocity of the rock when it hits the ground we need to plug in t = 5.376 into the velocity function from part a: V = 10 - 3.72(5.376) Then solve: V = -10m/s

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