A racecar is racing along a circular track. The car starts at the 3-o'clock position and travels CCW along the track. The car is constantly 8 feet from the center of the race track and travels at a constant speed and it takes the car 7.854 seconds to complete one full lap.

a. How many radians does the car sweep out per second? radians per second Preview

b. Write a function fthat determines the car's distance to the right of the center of the race track (in feet) in terms of the number of seconds t since the start of the race. f(t)

Answer:

a.  The car sweeps out 0.799 radians per second.

b. f(t) = 8 cos (0.799t)

Step-by-step explanation:

Given that:

The radius r = 8 feet

a.

The time t it takes to complete one full lap = 7.854 seconds

Recall that:

Angular velocity ω = 2π / t

So if in 7.854 seconds, the car sweeps = 2π radian

∴

per second, the car will sweep ω  = 2π/ 7.854

ω = 0.799 rad/sec

Thus, the car sweeps out 0.799 radians per second.

b.

Since, r = 8 feet.

Suppose the car travels an angle θ in time (t),

Then the general equation for the horizontal component is:

f(t) = rcos( ωt)

f(t) = 8 cos (0.799t)

The true statements are:

The given parameters are:

To do this, we simply calculate the angular velocity as follows:

[tex]\omega = \frac{2\pi}{T}[/tex]

Substitute 7.854 for T

[tex]\omega = \frac{2\pi}{7.854}[/tex]

Simplify

[tex]\omega = 0.255 \pi[/tex]

Hence, the car swept out [tex]0.255 \pi[/tex] radians per seconds

To do this, we simply make use of the general equation for the horizontal component as follows:

[tex]f(t) = r\cos( \omega t)[/tex]

Substitute 8 for r

[tex]f(t) = 8\cos( \omega t)[/tex]

Substitute [tex]\omega = 0.255 \pi[/tex]

[tex]f(t) = 8\cos( 0.255\pi t)[/tex]

Hence, the function f that determines the car's distance at time t is: [tex]f(t) = 8\cos( 0.255\pi t)[/tex]

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