1. 2.2b Rates of Change Position, velocity, acceleration

2. The position function (usually called s) relates position to time. You might be familiar with s(t) = -16t 2 + v 0 t + s 0 from physics or even algebra 2 This function is used when distance is measured in feet and v 0 is a fancy way of writing “velocity at time zero, or initial velocity” and s 0 is a fancy way of writing “position at time zero, or initial position”.

3. Let’s look at s(t) = -16t 2 - 30t + 150 Velocity to start is 30 ft/sec (someone pushed you!), you are falling from a cliff 150 feet high, and you want to know your position at different times. s(0) = ? s(1) = ? s(2) = ? How long till you splat on the ground? s(0) = 150 ft s(1) = -16 - 30 + 150 = 104 ft s(2) = -64 - 60 + 150 =36 ft Hitting the ground means position from the ground is 0, so solve t=2.26 or -4.14 seconds

5. If you want to find instantaneous velocity at a moment of time, you would want to calculate the average velocity as the change in time got smaller and smaller. In other words, find the rate of change (slope) of the position function at time t. Sounds like a derivative idea! The velocity function is the derivative of the position function! Speed of an object is the absolute value of velocity, because speed cannot be negative.

6. In general, the position function is given as where s 0 is initial position, v 0 is initial velocity, and g is the acceleration due to gravity. On earth that is -32 feet per second per second or -9.8 meters per second per second.

8. Notice the diver moves upward for the first half second, because the velocity is positive for the first half second. When the velocity is zero, the diver has reached the maximum height of the dive.

9. 2.2b Assignment: p. 115 #75-76, 83-90, 93-96, 103-104