Nahee is running around the circular track with equation x2+y2=8100 at a speed of 5.8 meters per second.

Question

Nahee is running around the circular track with equation x2+y2=8100 at a speed of 5.8 meters per second. She starts at the point (0,90) and runs counterclockwise. What are Nahee's coordinates after she has been running for 20 minutes?

Answer 1

There are several different ways to approach this problem and I'm unsure of the methods you were taught but I'll consider this with regards to circular motion. Here's a sketch:

For uniform circular motion, her angular velocity (ω) can be calculated as her tangential velocity (5.8 m/s) divided by the radius so ω = 5.8/90 = 0.064444 radians/sec

20 minutes is 20•60 seconds or 1200 seconds meaning she traveled 0.064444•1200 or 73.3333 radians. There are 2π radians in one revolution so we can subtract out increments of 2π to see where she is after 20 minutes. Since 73.3333/(2π) = 12.31 revolutions, lets subtract 24π (12•2π) from 73.3333:

73.33333 - 24π = 1,9351 radians. So she is 1.9351 radians past the top of the circle (about 110°, so just past the (-90,0) point).

We can find the x and y locations of that point by using the sine and cosine functions.

y = rsin(θ) and x = rcos(θ) HOWEVER, to get the actual angles from the standard starting location, we must add π/2 to the 1.9351 since we started at the top of the circle and the standard coordinate framework starts at the (90, 0) location. So θ in standard form is 1,9351 + π/2 = 3.5059 radians. So:

x = 90cos(3.5059 radians) = -84.093 meters

y = 90sin(3.5059 radians) = -32.068 meters

So using the center of the circle as the origin, she is 84.093 meters left of origin and 32.068 meters down from the origin or at the point (-84.093, -32.068)