Question

30 POINTS PLEASE HELP THIS IS MISSING FOR WEEKS THANK YOU

A scientist uses a submarine to study ocean life..
-She begins at sea level,which is elevation of 0 feet
-She travels straight down for 80 seconds at a speed of 4.1 feet per second.
-She then travels directly up for 50 secs at a speed of 2.9 feet per second
PART A
What is the location of the sumberine after the 120-second period

Answer= ________feet

PART B
After this 130-second period,how much time,in seconds,will it take for the scientist to travel back to sea level at the submarine's maximum speed of 5.6 feet per second?

Answer:____________Seconds

Answer 1

A.) 212ft

B.) 32.68 --> 32.7 seconds

Explanation:

A.)

We start with 0 as stated. She travels down for 80 seconds at 4.1ft per second

So we need to first calculate (80)(4.1). 80 seconds of diving 4.1ft/sec

80 * 4.1 = 328ft

She then travels up for 50 seconds at 2.9ft/sec

We'll do the same thing

50 * 2.9 = 145ft.. we'll save this for later

Part A wants 120sec period.

40 * 2.9 = 116ft

We'll subtract 328 - 116 to get 212ft

B.) Now we can use the 145 from earlier

328 - 145 = 183

So now we're going 5.6ft/sec until we hit 0

So we'll divide 183/5.6 to get 32.68 seconds or 32.7

Answer 2

PART A:

The location of the submarine after 80 seconds of traveling straight down at 4.1 feet per second is 80 x 4.1 = 332 feet below sea level.

After 50 seconds of traveling directly up at 2.9 feet per second, the submarine will be at 50 x 2.9 = 145 feet below sea level.

So the submarine's location after 120 seconds is 332 - 145 = 187 feet below sea level.

PART B:

To travel back to sea level, the submarine needs to go up 187 feet.

At the submarine's maximum speed of 5.6 feet per second, it would take 187 / 5.6 = 33.39 seconds to travel back to sea level.

Thus, the answer is 33 seconds.