Question

(3) Two people are at an elevator. At the same time one person starts to walk away from the elevator at a rate of 2 ft/sec and the other person starts going up in the elevator at a rate of 7 ft/sec. What rate is the distance between the two people changing 15 seconds later

Answer 1

Answer:

The rate at which the distance between the two people is changing 15 seconds later is 7.28 ft/sec.

Step-by-step explanation:

Here, we note the following

Let the person walking way fom the elevator be X

Let the other person going up in the elevator be Y

Therefore after 15 seconds, their positions will be;

For X,  2 ft/sec × 15 s = 30 ft away from the elevator

For Y,  7 ft/sec × 15 s = 105 ft up in the elevator

At that instant, the distance between them is given as

d² = x² + y²

= 30² +105² = 11925 ft²

d = √11925 ft²  = 109.202 ft

The rate of change of the distance between the two peopl, X and Y is given as

$\frac{dd^2}{dt} = \frac{dx^2}{dt} + \frac{dy^2}{dt}$

Therefore,  $2d\frac{dd}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ or $d\frac{dd}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}$

Since $\frac{dx}{dt}$ is given as 2 ft/sec and $\frac{dy}{dt}$ = 7 ft/sec

Then  $d\frac{dd}{dt} = 30 \times 2 + 105 \times 7 = 795$

$\frac{dd}{dt} = \frac{795}{109.202 } =7.28 ft/sec$

Answer 2

Answer:

The rate at which the distance between the two people change is 7.28 ft/s

Step-by-step explanation:

Since the first person started walking away from the elevator at a rate of 2ft/s. In 15 seconds, the distance that he/she must have covered is:

1 sec ------ 2ft

15 sec ------ ?ft

= 15/1 × 2

= 30 ft

On the other hand, if the other person started moving up in the elevator at a rate of 7ft/s, then in 15 seconds this person must have covered:

1 sec -------7ft

15 sec ------ ?ft

= 15/1 × 7

= 105ft

The shape formed as a result of the movement of these two people is a right - angle triangle.

Since the first person is moving away from the elevator at a rate of 2ft/s and the second person is going up in the elevator at a rate of 7ft/s, the rate of change of the distance between the two people can be calculated using the product rule:-

2m × dm/dt = [(2x × dx/dt ) + (2y × dy/dt)]

In this case "x" = the distance covered by the first person after 15 seconds (30ft)

dx/dt = the rate of change of the first person's distance (2ft/s)

y = the distance covered by the second person after 15 seconds (105ft)

dy/dt = the rate of change of the second person's distance (7ft/s)

m = the distance between these two people after 15 seconds.

dm/dt = the rate of change of the distance between the two people (what we are required to calculate)

It is obvious that we need "m" in order to calculate the rate at which the distance between the 2 people change after 15 seconds.

Since the shape created by their movement is a right angle, we can determine/calculate "x" using the Pythagoras theorem:

m^2 = A^2 + B^2

Where m = the distance between the two people after 15 seconds

A = distance covered by the first person after 15 seconds.

B = distance covered by the second person after 15 seconds

m^2 = 30^2 + 105^2

m^2 = 900 + 11025

m^2 = 11,925

m = √11,925

m = 109.2ft

Since the distance between them after 15 seconds is 109.2ft, we can then fix all these terms into the product rule and calculate the rate of change of distance between the two people.

2 × 109.2 × dm/dt = [(2 × 30 ×2) + (2 × 105 × 7)]

218.4 × dm/dt = [120 + 1,470]

218.4 × dm/dt = 1,590

dm/dt = 1,590/218.4

dm/dt = 7.28ft/s

Therefore, the rate of change of the distance between the two people is 7.28ft/s