Question

Erica is participating in a road race. The first part of the race is on a 5.2-mile-long straight road oriented at an angle of 25∘ north of east. The road then turns due north for another 5.0 mi to the finish line.

In miles, what is the straight-line distance from the starting point to the end of the race?

Answer 1

To find the straight-line distance from the starting point to the end of the race, we can use vector addition and the Pythagorean theorem. The east and north components of the displacement can be found using trigonometry, and then the straight-line distance can be calculated using the Pythagorean theorem.

Step-by-step explanation:

The straight-line distance from the starting point to the end of the race can be found using vector addition. First, we need to find the east and north components of the displacement. The first part of the race is on a 5.2-mile-long straight road oriented at an angle of 25∘ north of east. Using trigonometry, we can find the east component by multiplying the length of the road by the cosine of the angle. The north component can be found by multiplying the length of the road by the sine of the angle. Adding the east and north components, we get the displacement. To find the straight-line distance, we can use the Pythagorean theorem. The east and north components form the two sides of a right triangle, with the straight-line distance as the hypotenuse. Therefore, we can use the equation a^2 + b^2 = c^2, where a and b are the east and north components respectively, and c is the straight-line distance.

In this case, the east component is 5.2 miles * cos(25∘) = 4.693 miles, and the north component is 5.2 miles * sin(25∘) = 2.191 miles. Using the Pythagorean theorem, we have (4.693 miles)^2 + (2.191 miles)^2 = c^2. Solving for c, we get c = 5.237 miles.

Answer 2

8.6 miles

Step-by-step explanation:

We need to calculate the components of the total displacement along the east-west and north-south directions first.

In the first part, Erica moves 5.2 miles at 25∘ north of east. So the components of this displacement along the two directions are:

East:
`d_(1x) = 5.2 cos 25^(\circ)=4.7 mi`

North:
`d_(1y) = 5.2 sin 25^(\circ)=2.2 mi`

In the second part, Erica moves 5.0 miles north. So, the components of this displacement are:

East:
`d_(2x)=0`

North:
`d_(2y) = 5.0 mi`

So the components of the total displacement are

East:
`d_x = d_(1x)+d_(2x)=4.7 + 0 = 4.7 mi`

North:
`d_y = d_(1y)+d_(2y)= 2.2 + 5.0 = 7.2 mi`

Therefore the magnitude of the displacement, which is the straight-line distance from the starting point to the end of the race, is

`d=√(d_x^2 +d_y^2)=√(4.7^2+7.2^2)=8.6 mi`