Question

When an object is dropped straight down, the distance (in feet) that it travels in t seconds is given by s(t) = -16t²

Find the velocity at each of the following times.
The velocity after 4 seconds is ______ ft/sec
The velocity after 5 seconds is ______ ft/sec
The velocity after 8 seconds is ______ ft/sec
The acceleration is _______ ft/sec

Answer 1

The velocity of an object dropped straight down can be found by taking the derivative of the position function with respect to time. The velocity after 4 seconds is -128 ft/sec, after 5 seconds is -160 ft/sec, and after 8 seconds is -256 ft/sec. The acceleration of the object is constant at -32 ft/sec².

Step-by-step explanation:

The distance traveled by an object dropped straight down is given by the equation s(t) = -16t², where t is the time in seconds. To find the velocity at a given time, we take the derivative of the position function with respect to time. The derivative of -16t² is -32t. Therefore, the velocity at any time t is -32t ft/sec.

Calculating the velocity:

1. Velocity after 4 seconds: -32 * 4 = -128 ft/sec
2. Velocity after 5 seconds: -32 * 5 = -160 ft/sec
3. Velocity after 8 seconds: -32 * 8 = -256 ft/sec

The acceleration of the object is constant and equal to -32 ft/sec². Since the object is accelerating downward, the acceleration is negative.