Question

Distance Between Two Cars A car leaves an intersection traveling west. Its position 4 sec later is 19 ft from the intersection. At the same time, another car leaves the same intersection heading north so that its position 4 sec later is 26 ft from the intersection. If the speeds of the cars at that instant of time are 8 ft/sec and 12 ft/sec, respectively, find the rate at which the distance between the two cars is changing. (Round your answer to one decimal place.)

Answer 1

The answer to the question is;

The rate at which the distance between the two cars is changing is equal to 14.4 ft/sec.

Step-by-step explanation:

We note that the distance traveled by each car after 4 seconds is

Car A = 19 ft in the west direction.

Car B = 26 ft in the north direction

The distance between the two cars is given by the length of the hypotenuse side of a right angled triangle with the north being the y coordinate and the west being the x coordinate.

Therefore, let the distance between the two cars be s

we have

s² = x² + y²

= (19 ft)² + (26 ft)² = 1037 ft²

s =
`√(1037 ft^2)` = 32.202 ft.

The rate of change of the distance from their location 4 seconds after they commenced their journeys is given by;

Since s² = x² + y² we have

`(ds^(2) )/(dt) = (dx^(2) )/(dt) + (dy^(2) )/(dt)`

→
`2s(ds )/(dt) = 2x(dx)/(dt) + 2y(dy )/(dt)` which gives

`s(ds )/(dt) = x(dx)/(dt) + y(dy )/(dt)`

We note that the speeds of the cars were given as

Car B moving north = 12 ft/sec, which is the y direction and

Car A moving west = 8 ft/sec which is the x direction.

Therefore

`(dy )/(dt)` = 12 ft/sec and

`(dx)/(dt)` = 8 ft/sec

`s(ds )/(dt) = x(dx)/(dt) + y(dy )/(dt)` becomes

`32.202 ft.(ds )/(dt) = 19 ft * 8 (ft)/(sec) + 26ft* 12(ft)/(sec)` = 464 ft²/sec

`(ds )/(dt) = (464(ft^(2) )/(sec) )/(32.202 ft.)` = 14.409 ft/sec ≈ 14.4 ft/sec to one place of decimal.