Question

A motorcycle travels 20 miles west and then turns and goes 52° north of west another 38 miles. How far is the motorcycle from its starting point?? Round your answer to the nearest tenth

__ miles

Answer 1

The motorcycle is approximately 25.7 miles from its starting point. The total distance is calculated using vector addition and the Pythagorean theorem.

Step-by-step explanation:

To find the distance from the starting point, we can use the concept of vector addition.

First, we need to find the x-component and y-component of the motorcycle's displacement.

For the 20 miles west, the x-component is -20 miles (negative because it is in the west direction) and the y-component is 0 miles (no displacement in the north direction).

For the 38 miles at 52° north of west, we can use trigonometry to find the x-component and y-component. The x-component is 38 miles * sin(52°) = 29.2 miles (negative because it is in the west direction) and the y-component is 38 miles * cos(52°) = 24.3 miles (positive because it is in the north direction).

Now, we can find the total displacement by adding the x-components and y-components separately.

x-component = -20 miles + 29.2 miles = 9.2 miles

y-component = 0 miles + 24.3 miles = 24.3 miles

Using the Pythagorean theorem, we can find the distance from the starting point:

distance = sqrt((9.2 miles)^2 + (24.3 miles)^2) = 25.7 miles (rounded to the nearest tenth).

Answer 2

52.7 miles

Step-by-step explanation:

Please refer to attached photo. (Apologies for the terrible drawing).

We can deduce from A to B the distance is 20 miles, therefore no calculations is required.

However, from B to C, we will have to find the horizontal and vertical distances.

Horizontal Distance B > C = = hBC =
`BCcos(52)` =
`38cos(52) miles`

Vertical Distance B > C = vBC =
`BCsin(52)` =
`38sin(52)miles`

Here you can see the whole diagram is a right angle triangle. Which means, we can use Pythagoras' Theorem to find AC.

By Pythagoras Theorem,

`c^(2) =a^(2)+b^(2)`

`AC^(2) = (AB + hBC)^(2) + vBC^(2) \\AC^(2) = (20+38cos(52))^(2) +(38sin(52))^(2) \\AC = \sqrt{(20+38cos(52))^(2) +(38sin(52))^(2) } \\AC = 52.7 miles`(nearest tenth)