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Calculus BC: Applications of the Derivative

Problems

Velocity and Acceleration

Analysis of Graphs

Problem : Suppose a rock is thrown straight up from atop a 200 -meter-high cliff at an initial speed of 30 feet per second. The height, in meters, of the rock above the ground (until it lands) at time t is given by the function h(t) = - gt 2/2 + 30t + 200 , where g 9.81 is a constant of gravitational acceleration. When does the rock reach its maximum height? What is this maximum height? How fast is the rock moving after 3 seconds?

When the rock reaches its maximum height, it is instantaneously stationary, with speed 0 . Solving

h'(t) = - gt + 30 = 0    

for t , we obtain t = 30/g 3.06 as the time when the rock reaches its maximum height. Substituting back into h(t) , we find that the maximum height is

h(30/g) = +30 +200 = +200 245.89    

measured in meters. To find the speed at time t = 3 , we compute

h'(3) = (- g)(3) + 30 0.58    

meters per second, which makes sense, because the rock is about 0.06 seconds away from reaching its maximum height and coming to an instantaneous stop.

Problem : The position of a box, in a certain coordinate system, attached to the end of a spring is given by p(t) = sin(2t) . What is the acceleration of the box at time t ? How does this relate to its position?

The velocity of the box is equal to

p'(t) = 2 cos(2t)    

and the acceleration is given by

p''(t) = - 4 sin(2t) = - 4p(t)    

This makes sense, because the spring should exert a restoring force proportional to the displacement of the box and in the opposite direction from the displacement.

Problem : Suppose the velocity of a sprinter (in meters per second) at time t seconds after the start of a 40 meter dash is given by

v(t) = 3 log(t + 1)    

How fast is the sprinter accelerating 1 second after she starts sprinting?

The acceleration is given by

v'(t) =    

at time t . Thus the acceleration at time t = 1 is 3/2 meters per second per second.

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