If a rock is thrown upward on the planet Mars with a velocity of 15 m/s, its height above the ground (in meters) after t seconds is given by H = 15t − 1.86t2.(a)Find the velocity (in m/s) of the rock after 2 seconds. m/s(b)Find the velocity (in m/s) of the rock when t = a.15−3.72a m/s(c)When (in seconds) will the rock hit the surface? (Round your answer to one decimal place.)t = s(d)With what velocity (in m/s) will the rock hit the surface? m/s
Question
If a rock is thrown upward on the planet Mars with a velocity of 15 m/s, its height above the ground (in meters) after t seconds is given by H = 15t − 1.86t2.(a)Find the velocity (in m/s) of the rock after 2 seconds. m/s(b)Find the velocity (in m/s) of the rock when t = a.15−3.72a m/s(c)When (in seconds) will the rock hit the surface? (Round your answer to one decimal place.)t = s(d)With what velocity (in m/s) will the rock hit the surface? m/s
Solution
(a) The velocity of the rock at any time t is given by the derivative of the height function H(t). So, we need to find the derivative of H = 15t - 1.86t^2. The derivative is H' = 15 - 3.72t. To find the velocity after 2 seconds, we substitute t = 2 into the derivative to get H'(2) = 15 - 3.72*2 = 7.56 m/s.
(b) The velocity of the rock at any time t = a is given by H'(a) = 15 - 3.72a m/s.
(c) The rock hits the surface when H(t) = 0. So, we need to solve the equation 15t - 1.86t^2 = 0 for t. This is a quadratic equation, and its solutions are given by the quadratic formula t = [-b ± sqrt(b^2 - 4ac)] / (2a), where a = -1.86, b = 15, and c = 0. Solving this equation gives t = 8.1 seconds (rounded to one decimal place).
(d) The velocity of the rock when it hits the surface is given by H'(t) at t = 8.1. Substituting t = 8.1 into the derivative gives H'(8.1) = 15 - 3.72*8.1 = -15 m/s. The negative sign indicates that the rock is moving downward.
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