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) = - gt2/2 + 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
at time t. Thus the acceleration at time t = 1 is 3/2 meters per second per second.