Remember that Isaac Newton was trying to understand planetary motion when he discovered the fundamentals of differential calculus. He was studying the relationships of distance, velocity, time, and acceleration.
We will now work a simple motion problem using the formula described at right which is based on Newtonian physics. If a ball is released from a height of 5 feet with an initial velocity upward of 64 feet per second, we would like to determine:
1. What is the average velocity from .5 to 1 second?
2. What is the instantaneous velocity at 1 second?
1. Average velocity is the change in distance divided by the increment of time during which the distance changed. Substituting the values given above into our formula at right, we get,
s(t) = 5 + 64t -16t2. From .5 to 1 second, the change in time is t2 - t1 = 1 - .5 = .5 seconds. Over this time period, the change in distance is s(1) - s(.5) = 53-33 = 20 feet. Then, the average velocity is the change in distance divided by the change in time, or 20/.5, which equals 40. The units are distance divided by time; feet per second in our example. We conclude that the average velocity from .5 to 1 second after throwing the ball upwards at 64 feet per second, is 40 feet per second. Remember that gravity acts against the upward motion of the ball, so the ball slows down at first.
2. Instantaneous velocity is the derivative of distance with respect to time. To compute the derivative of the distance function, we first compute the difference quotient and simplify it using algebra:
. Then we take the limit,
. So, the formula for the derivative is s'(t)=64-32t. and the instantaneous velocity at 1 second is s'(1)=64-(32)(1)=32. The units for the derivative are feet per second, the units of s divided by t. This instantaneous velocity at 1 second is less than the average velocity from .5 to 1 second that we computed in the first part of this problem because the ball is slowing down under the influence of gravity from .5 to 1 second. (Note that this solution value of 32 could also be interpreted from the graphical perspective; the slope of the tangent line to the graph of s(t) at t=1 is 32.)
A ball is thrown directly upward. After some period of time, the ball reaches its maximum height and then starts to fall. A simple formula that describes this motion is, s(t) = s0 + v0t -16t2, where s0 is the initial height in feet, v0 is the initial velocity (a positive number means the ball is moving upward) in units of feet per second, and t is the time in seconds since the ball was thrown. The number 16 is half of the acceleration of gravity near the earth surface, 32 feet per second per second. Gravity pulls the ball downward.
