What is the distance between the two cars?

Question

Two cars are driving in opposite direction and travel 4 miles each; then, both make a left turn and travel 3 more miles.

Answer 1

10 miles

Each car is driving the same distance in opposite directions from their point of origin. Think of it as driving 4 miles on the x-axis of a graph, then 3 miles on the y-axis of a graph. If you draw it, you would have 2 right triangles - one in the positive-x positive-y quadrant, and the other in the negative-x negative-y quadrant.

For each car, the distance from the origin is then
sqrt(4^2 + 3^2) = sqrt(16 + 9) = sqrt(25) = 5

Since this is the distance from the origin of each car, their separation is then 5 + 5 = 10 miles.

Answer 2

The cars will then be 10 miles apart. Each car has driven the equivalent of "two legs of a 3,4,5 right triangle" if you think about it that way. And distance between them is essentially "the sum of the triangle's hypotenuses" in this case ... twice 5 makes 10 miles.

Hope this metaphor helps!

Answer 3

10 miles apart.

This is an example of the classic 3-4-5 right triangle.

Answer 4

10 miles

Answer 5

10 miles because you have a right triangle with sides 4x2 = 8 miles and 3x2 = 6 miles, so the hypothenuse is sqrt(8^2 + 6^2) = sqrt(64+36) = sqrt(100) = 10

Answer 6

10 miles
To see this set up right triangles from the point of initial separations. The sides of these right triangles will be 4 and 3, therefore, each hypotenuse will be 5 miles long.

Answer 7

the fee of substitute of the gap between the two vehicles is easily the RELATIVE velocity between those 2 vehicles. So, in case you will get that, you get the answer. Now, we could calculate the relative velocity of the west-vehicle with appreciate to the south-vehicle. velocity of west-vehicle, Vw= i. [25 mi/h] velocity of south-vehicle, Vs= j. [60 mi/h], right here i, j are basically 2 unit vector, there beneficial course are in west and south respectively. Now, the relative velocity of the west-vehicle with appreciate to the south-vehicle, Vws = Vw - Vs = i. [25 mi/h]- j. [60 mi/h], that's cost is sq. root of (25^2 + 60^2) = sixty 5 mi/h, that's the answer. As you will discover the answer would not have any time correct term, for this reason it is not correct whether after 2 hrs or 20 hrs, the fee would be comparable constantly as long as they maintain their uniform velocity.