Mathematics: 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.

19095705.3333 is the answer hope i helped!...

A car leaves an intersection traveling west. Its

KarateKat

21.01.2022

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.

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