Question

If a rock is thrown upward on the planet mars with a velocity of 11 m/s, its height (in meters) after t seconds is given by h = 11t − 1.86t2. (a) find the velocity of the rock after two seconds. 3.56 m/s (b) find the velocity of the rock when t =
a. 11−3.72a m/s (c) when will the rock hit the surface? (round your answer to one decimal place.) t = s (d) with what velocity will the rock hit the surface? m/s

Answer 1

(a) 3.56 m/s 
(b) 11 - 3.72a 
(c) t = 5.9 s 
(d) -11 m/s  
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule. 
y = 11t - 1.86t^2 
y' = 11 - 3.72t 
Now that you have the first derivative, it will give you the velocity as a function of t. 
(a) Velocity after 2 seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56 
So the velocity is 3.56 m/s  
(b) Velocity after a seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72a  
So the answer is 11 - 3.72a  
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.  
(d) Plug in the value of t calculated for (c) into the velocity function, so: 
y' = 11 - 3.72a
 y' = 11 - 3.72*5.913978495
 y' = 11 - 22
 y' = -11 
 So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.