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3 years ago

12

Car A is traveling north on a straight highway and car B is traveling west on a different straight highway. Each car is approach

ing the intersection of these highways. At a certain moment, car A is 0.4 km from the intersection and traveling at 75 km/h while car B is 0.3 km from the intersection and traveling at 70 km/h. How fast is the distance between the cars changing at that moment? km/h

1 answer:

Inessa05 [86]3 years ago

7 0

Answer:

102 km/h

Explanation:

x = distance of car A at any time from the intersection

x₀ = distance of car A at some time = 0.4 km

$v_{A}$ = Speed of car A = 75 km/h

y = distance of car B at any time from the intersection

y₀ = distance of car B at some time = 0.3 km

$v_{B}$ = Speed of car B = 70 km/h

d = distance between the two cars at any time

d₀ = distance between the two cars at some time

v = rate of change of distance between the cars

Using Pythagorean theorem

d²₀ = x₀² + y₀²

d²₀ = 0.4² + 0.3²

d₀ = 0.5 m

Distance between the two cars at any time is given using Pythagorean theorem as

d² = x² + y²

Taking derivative both side relative to "t"

$2d \left ( \frac{dd}{dt} \right ) = 2x ( \frac{dx}{dt} \right ) + 2y ( \frac{dy}{dt} \right )$

$d_{o} v = x_{o} ( \frac{dx}{dt} \right ) + y_{o} ( \frac{dy}{dt} \right )$

(0.5) v = (0.4) $v_{A}$ + (0.3) $v_{B}$

(0.5) v = (0.4) (75) + (0.3) (70)

v = 102 km/h

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Explanation:

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What is the current I(3τ), that is, the current after three time constants have passed? The current in the circuit will approach

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$I(\tau)=0.051 A$

b

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c

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Explanation:

From the question we are told that

$I(t) = \frac{e}{R}(1-e^{\frac{t}{\tau} }) ; \ Where \ \tau = L/R$

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