Question

Two cars leave an intersection at the same time. One is headed south at a constant speed of 40 miles per hour, the other is headed west at a constant speed of 30 miles per hour (see the figure). Express the distance d between the cars as a function of the time t. (Hint: At t = 0 the cars leave the intersection.)

d(t)= ...?

Answer 1

The distance between the two cars can be expressed as a function of time using the Pythagorean theorem. At t = 0, the cars are 2 km apart, and as time progresses, their distances increase.

Step-by-step explanation:

The distance between the two cars can be expressed as a function of time using the Pythagorean theorem. Let's consider the time t as the independent variable. At t = 0, both cars leave the intersection, so the distance between them is initially given by:

`d(0) = √(((2 km)^2 + (0 km)^2)) = √((4 km^2)) = 2 km`

As time progresses, the car headed south travels at a speed of 40 mph, which means its distance from the starting point increases by 40t miles. Similarly, the car headed west travels at a speed of 30 mph, increasing its distance from the starting point by 30t miles.

Using the Pythagorean theorem again, we can find the distance d between the two cars as a function of time:

`d(t) = \sqrt{((40t)^2 + (30t)^2)`

Answer 2

I got into a mess of trouble when I reached the part where it says
"(see the figure)". But I think I was able to get enough out of the
rest of the question to answer it.

One car is headed south, and the other car is headed west.
So the cars are driving on the legs of a right triangle, and the
hypotenuse is always the line between the cars.

First car: Distance from the starting point after 't' hours = 40 t miles.

Second car: Distance from the starting point after 't' hours = 30 t miles.

Distance between the cars

= hypotenuse of the right triangle

= √(one leg² + other leg²)

= √[ (40t miles)² + (30t miles)² ]

= √ (1600t² miles² + 900t² miles²)

= √ 2500 t² miles²

d(t) = 50 t miles .

The cars are 50 miles apart after 1 hour, 100 miles apart after 2 hours,
150 miles after 3 hours, 200 miles after 4 hours, . . . , etc.