A scientist travels straight down for 90 seconds at a speed of 3.5 feet per second and then travels directly up for 30 seconds at a speed at a speed of 2.2 feet per second after this 120 second period,how much time in seconds,will it take for the scientist to travel back to sea level at maximum speed of 4.8 feet per second?(round answer to the nearest tenth of a second).
d = d1 - d2 = 3.5*90 - 2.2*30 = 315 - 66 = 249 ft. down.
d = V*t = 249.
4.8 * t = 249,
t =
To solve this problem, we need to find the time it will take for the scientist to reach sea level again.
First, let's determine the distance traveled during the 90 seconds of downward travel:
Distance = Speed x Time
Distance = 3.5 feet/second x 90 seconds
Distance = 315 feet
Next, let's determine the distance traveled during the 30 seconds of upward travel:
Distance = Speed x Time
Distance = 2.2 feet/second x 30 seconds
Distance = 66 feet
Since the scientist traveled both upward and downward, the net distance traveled from the starting point is:
Net Distance = Distance Downward - Distance Upward
Net Distance = 315 feet - 66 feet
Net Distance = 249 feet
Now, let's calculate the time it will take for the scientist to reach sea level again at a maximum speed of 4.8 feet per second:
Time = Distance / Speed
Time = 249 feet / 4.8 feet/second
Time ≈ 51.875 seconds
Therefore, it will take approximately 51.9 seconds for the scientist to travel back to sea level at a maximum speed of 4.8 feet per second.