Question

PROJECTILE An object is launched into the air and then falls to the ground. Its velocity is modeled by the equation v = 200- 321,

where the velocity v is measured in feet per second and time t is measured in seconds. The object's speed is the absolute value of its
velocity. Write and solve a compound inequality to determine the time intervals in which the speed of the object will be between 40 and
88 feet per second. Interpret your solution in the context of the situation.
40 1200 321 <
3.5 < 1<
-
The speed is between 40 and
going up and from 7.5 to
or 7.5 < t <
feet per second in the intervals from
seconds coming down.
to 5 seconds

Answer 1

To determine the time intervals in which the speed of the object will be between 40 and 88 feet per second, we need to consider the velocity equation v = 200 - 321t and find the range of values for t.

First, let's set up the compound inequality for the speed:

40 < |200 - 321t| < 88

Next, let's solve this compound inequality:

40 < 200 - 321t < 88

Subtract 200 from all three parts:

-160 < -321t < -112

Divide all three parts by -321 (note that dividing by a negative number flips the direction of the inequality):

160/321 > t > 112/321

Simplifying:

0.498 < t < 0.349

So, the time intervals in which the speed of the object will be between 40 and 88 feet per second are 0.349 < t < 0.498 seconds.

Interpretation:

During the time interval from approximately 0.349 to 0.498 seconds, the speed of the object will be between 40 and 88 feet per second. This corresponds to a specific portion of the object's trajectory, either during its upward or downward motion.