Question

A car leaves an intersection traveling west. Its position 4 sec later is 18 ft from the intersection. At the same time, another car leaves the same intersection heading north so that its position 4 sec later is 27 ft from the intersection. If the speeds of the cars at that instant of time are 7 ft/sec and 14 ft/sec, respectively, find the rate at which the distance between the two cars is changing.

Answer 1

Answer:

The rate at which the distance between the two cars is changing is;

15.53 ft/sec.

Step-by-step explanation:

To solve the question, we note that

Position of car A 4 s after start of motion, w = 18 ft west,

Position of car B 4 s after start of motion, n = 27 ft north

Therefore

The distance between the two cars at the 4 s instance is

d² = w² + n²

d² = 18² + 27² = 1053 ft² and

d = 32.45 ft

The rate at which the distance between the two cars is changing is given by;

Differentiating both sides of the equation, d² = w² + n², with respect to t as follows.

$\frac{dd^2}{dt} = \frac{dw^2}{dt} + \frac{dn^2}{dt} \Longrightarrow 2 d\frac{dd}{dt} = 2w\frac{dw}{dt} + 2n\frac{dn}{dt}$

It is given that the speeds of car A and car B at the 4 second instant are 7 ft/sec and 14 ft/sec, respectively

That is;

$\frac{dw}{dt} = 7\frac{ft}{sec}$ and $\frac{dn}{dt} = 14\frac{ft}{sec}$

Substituting the values of speed in the equation of rate of change gives

$2 d\frac{dd}{dt} = 2w\frac{dw}{dt} + 2n\frac{dn}{dt}\Longrightarrow d\frac{dd}{dt} = w\frac{dw}{dt} + n\frac{dn}{dt}$

$d\frac{dd}{dt} = w\frac{dw}{dt} + n\frac{dn}{dt} \Longrightarrow 32.45\frac{dd}{dt} = 18\times 7 + 27\times 14 = 504$

$32.45\frac{dd}{dt} = 504$

$\frac{dd}{dt} = \frac{504}{32.45} = 15.53 \frac{ft}{sec}$

The rate at which the distance between the two cars is changing = dd/dt = 15.53 ft/sec.

Answer 2

Answer:

The rate of change of distance between the two cars is 15.53 ft/sec

Step-by-step explanation:

The first car, leaving the intersection travelled due west and the second car leaving the intersection travelled due north; we can imagine that the situation or shape created by the respective journeys of these vehicles is a right - angle triangle.

In order to calculate the rate of change of distance between the two cars, we need to apply the knowledge of product rule. Before then, let's call the distance that the first car has currently covered, "A", the distance that the second car covered, "B" and the current distance between the two cars, "C".

We can also denote the rate of change of the first car's distance by dA/dt and denote the rate of change of the second car's distance by dB/dt. Let the rate of change between the two cars be dC/dt.

The product rule =

2 × C × dC/dt = [(2 × A × dA/dt) + (2 × B × dB/dt)]

Where A = 18ft

B = 27ft

dA/dt = 7ft/sec

dB/dt = 14ft/sec

dC/dt = The unknown variable to be calculated (rate of change of distance between the two cars).

Before we can calculate dB/dt, we need to know the current distance between the two cars, "C". If the cars from the point of intersection travelled due west and due north respectively, then we can determine the current distance between them after they both travelled for 4 seconds. Since the shape formed by the cars is a right angle triangle, then we can calculate their current distance using Pythagoras theorem.

C^2 = A^2 + B^2

Where A = 18ft and B = 27 feet.

C^2 = 18^2 + 27^2

C^2 = 324 + 729

C^2 = 1,053

C = √1053

C = 32.45 feet

Since, we now know the current distance between the two vehicles after 4 seconds, all the variables needed for the calculation of dC/dt are now complete.

Fixing them in the product rule, we have:-

2 × 32.45 × dC/dt = [(2 × 18 × 7) + (2 × 27 × 14)]

64.9 × dC/dt = (252 + 756)

64.9 × dC/dt = 1008

dC/dt = 1008/64.9

dC/dt = 15.53 ft/sec

Therefore the rate of change of distance between the two cars is 15.53 ft/sec