SOLUTION: a car travels from point C northward for 1 hr then eastward for half an hour, then shifted N 30 E. if the constant speed of the car is 40kph, how far directly from point c will it

Algebra ->  Length-and-distance -> SOLUTION: a car travels from point C northward for 1 hr then eastward for half an hour, then shifted N 30 E. if the constant speed of the car is 40kph, how far directly from point c will it       Log On


   



Question 1202456: a car travels from point C northward for 1 hr then eastward for half an hour, then shifted N 30 E. if the constant speed of the car is 40kph, how far directly from point c will it be after 2 hrs.
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: 64.6965 km (approximate)


Explanation:

Draw an xy axis.
Place point C at the origin.

The car drives north at 40 kph for 1 hour.
Distance = rate*time
Distance = (40 kph)*(1 hr)
Distance = 40 km
The car drives 40 km north for the first leg of the trip.
Move from point C(0,0) to point A(0,40)

For the next leg of the trip, we have
Distance = rate*time
Distance = (40 kph)*(1/2 hr)
Distance = 20 km
This will move us from A(0,40) to B(20,40) when going 20 km east.

The notation N 30 E means "face north, then turn 30 degrees eastward".
This is equivalent to the notation E 60 N where we face east and then turn 60 degrees toward the north.
30+60 = 90
See this page for more info about compass bearings.
http://academic.brooklyn.cuny.edu/geology/leveson/core/linksa/comp.html

The notation E 60 N is more useful because 60 degrees is the reference angle.
Use polar coordinates as an offset to move from B(20,40) to D(30,57.320508)

The scratch work to determine the x,y coordinates of D are shown in this paragraph
r = distance traveled for the remaining 1/2 hr = 20 km
theta = 60 degree reference angle
(xB,yB) = x and y coordinates of point B
x = xB + r*cos(theta)
x = 20 + 20*cos(60)
x = 30
y = yB + r*sin(theta)
y = 40 + 20*sin(60)
y = 57.320508 which is approximate
Make sure your calculator is in degree mode.

To recap the journey:
  • Start at C(0,0)
  • Move 40 km north to arrive at A(0,40)
  • Move 20 km east to get to B(20,40)
  • Move 20 km along the compass bearing N 30 E (aka E 60 N) to arrive at the approximate location D(30,57.320508)
The pathway is C to A to B to D in that exact order.

The question is: how far is it from the start point C(0,0) to the end point D(30,57.320508)?

We could draw a right triangle and use the pythagorean theorem.
Or we could use the distance formula.

I'll use the distance formula.
(x1,y1) = (0,0) and (x2,y2) = (30,57.320508)
d+=+sqrt%28+%28x1-x2%29%5E2+%2B+%28y1-y2%29%5E2+%29
d+=+sqrt%28+%280-30%29%5E2+%2B+%280-57.320508%29%5E2+%29
d+=+sqrt%28+%28-30%29%5E2+%2B+%28-57.320508%29%5E2+%29
d+=+sqrt%28+900+%2B+3285.64063737806+%29
d+=+64.6965272435706
d+=+64.6965
This is the approximate distance from C(0,0) to D(30,57.320508)