Final answer:
To calculate the total distance traveled by the projectile in the first 5 seconds, one should compute the upward distance to the maximum height using integration of the velocity function, then compute the distance it falls down by integrating from the time it reaches the maximum height to 5 seconds, and sum the two distances.
Step-by-step explanation:
The question asks about calculating the total distance a projectile travels when launched straight up with an initial velocity of 100 ft/s and subjected to Earth's gravity. To determine the total distance traveled by the projectile during the first 5 seconds, we must consider the upward and downward paths. The velocity function is given by v(t) = 100 - 32t ft/s, which accounts for the acceleration due to gravity (32 ft/s2 downward).
First, we need to find the time at which the projectile reaches its maximum height. This occurs when its velocity is zero. Setting the velocity function equal to zero, we get tmax = 100/32 seconds. Since this time is less than 5 seconds, we know that the projectile reaches its maximum height and starts falling back down within the 5-second interval.
To find the distance traveled to the maximum height, we integrate the velocity function from 0 to tmax. The integral of v(t) from 0 to tmax is (100t - 16t2) evaluated from 0 to 100/32, which gives us the height H reached by the projectile. To find the total distance traveled in 5 seconds, we then calculate the distance fallen back down from tmax to 5 seconds by integrating the velocity function again but with the proper limits of integration.
After performing these calculations, we sum the upward and downward distances to compute the total distance traveled by the projectile during the first 5 seconds.