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. (Round your answer to one decimal place.)

Explanation:

Let x (t) is the distance of first car and the intersection at time t.

y(t) is the distance of second car and the intersection at time t.

z(t) is the distance between two cars at time t.

So,

Differentiate both sides wit respect to t.

2zz' = 2xx' = 2yy'

z' = (xx' + yy') / z .... (1)

now

x(t = 4s) = 18 ft

y(t = 4s) = 27 ft

x' (t = 4 s) = 7 ft/s

y' (t = 4s) = 14 ft/s

So,

z = 32.5 ft

So, z' = (18 x 7 + 27 x 14) / 32.5

z' = 15.5 ft/s

Thus, the rate of change of distance between two cars is 15.5 ft/s.

answer: he got 23

explanation:

trig 15.3 cos