A road perpendicular to a highway leads to a farmhouse located 2 km away. A car travels pastthe farmhouse on on the highway at a speed of 80 km/h. How fast is the distance between thecar and the farmhouse increasing when the car is 6 km past the intersection of the highwayand the road

In this exercise we want to calculate the speed of the vehicle to reach the farm, in this way we will find a speed of approximately:

To start this exercise we have to use some data informed in the text, like this:

Since distance is involved, time is required. Hence, we differentiate the equation above in respect to time

Calculate the speed at which the distance between the car and the farmhouse kept increasing when the car is 6 km past the intersection of the highway and the road. This calculated using the obtained differentiation above:

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75.9 km/hr

Step-by-step explanation:

Distance between the highway and farmhouse is given as = 2km = a

The distance after the intersection and the highway = b

Let the distance between the farmhouse and the car = c

Using the Pythagoras Theorem rule

c² = a² + b²

c² = 2² + b²

Step 1

Since distance is involved, time is required. Hence, we differentiate the equation above in respect to time

c² = 2² + b²

dc/dt (2c) = 4 + 2b

dc/dt =[ b/(√b² + 4)] × db/dt

We are told in the question that:

the car travels past the farmhouse on on the highway at a speed of 80 km/h.

We are asked to calculate the speed at which the distance between the car and the farmhouse kept increasing when the car is 6 km past the intersection of the highway and the road.

This calculated using the obtained differentiation above:

dc/dt = [ b/(√b² + 4)] × db/dt

Where b = 6km

db/dt = 80km/hr

[6/(√6² + 4)] × 80km/hr

6/√36 + 4 × 80km/hr

6 × 80/√40

480/√40

= 75.894663844km/hr

Approximately = 75.9km/hr

192/15 = 12.8.

12.8

step-by-step explanation:

192/15 is 12.8.

however, it says to round to the nearest foot, which would equal 13.