Question

1. a car is traveling around a circular racetrack with a radius of 120 meters. the car travels through a turn that measures 135° of the track’s circumference.

a) how far does the car travel during this turn?
b) if the turn covers a sector of the track (including the infield), what is the area of that sector?
round both answers to the nearest meter/square meter

Explanation:

Part (a)

Step1: Recall arc length formula

The formula for the length of an arc (which is the distance traveled along the sector's arc) is \( s = r\theta \), where \( \theta \) is in radians. First, convert \( 135^\circ \) to radians. We know that \( 1^\circ=\frac{\pi}{180} \) radians, so \( 135^\circ = 135\times\frac{\pi}{180}=\frac{3\pi}{4} \) radians. The radius \( r = 120 \) meters.

Step2: Calculate arc length

Substitute \( r = 120 \) and \( \theta=\frac{3\pi}{4} \) into the arc length formula: \( s=120\times\frac{3\pi}{4} \). Simplify this: \( 120\div4 = 30 \), so \( 30\times3\pi=90\pi \approx 90\times3.1416 = 282.744 \). Rounding to the nearest meter, we get \( 283 \) meters.

Step1: Recall sector area formula

The formula for the area of a sector is \( A=\frac{1}{2}r^2\theta \), where \( \theta \) is in radians. We already know \( r = 120 \) meters and \( \theta=\frac{3\pi}{4} \) radians.

Step2: Calculate sector area

Substitute the values into the formula: \( A=\frac{1}{2}\times(120)^2\times\frac{3\pi}{4} \). First, calculate \( (120)^2 = 14400 \). Then, \( \frac{1}{2}\times14400 = 7200 \). Next, \( 7200\times\frac{3\pi}{4}= 5400\pi \approx 5400\times3.1416 = 16964.64 \). Rounding to the nearest square meter, we get \( 16965 \) square meters.

Answer:

The car travels approximately \(\boxed{283}\) meters during the turn.

Part (b)