A race-car is driving counter-clockwise on a circular track with a radius of 1.9 miles. The car starts at the 3 o'clock position and travels at a constant speed of 85.5 miles per hour.

What distance (in miles) has the race-car traveled if the car has swept out an angle of 212 degrees?
_miles

What is the measure of the angle swept out by the car (in radians) if the car has traveled 4.9 miles?
_radians

Define a function, h, that gives the race-car's distance above the horizontal diameter of the track (in miles) in terms of the number of hours since the race-car started driving, t.

The car is moving in a circular track with a radius of 1.9 miles. The distance covered by the car if the track is revolved once = 2π * radius of the track.

a) Since the car has swept out an angle of 212 degrees, the distance covered by the race car = [tex]\frac{212}{360}*2\pi(1.9)=7.03 \ miles[/tex]

b) If the car traveled 4.9 miles, the angle swept out (θ) is:

[tex]4.9= \frac{\theta}{360} * 2\pi r\\\\\theta=\frac{4.9}{2\pi (1.9)}*360 \\\\\theta=147.76^o=147.76^o*\frac{\pi}{180} \\\\\theta=2.58\ rad[/tex]

c) h = distance covered, t = time in hourss.

Hence:

h = 85.5t

oeerivona oeerivona · Answer 2 · 2021-10-30T13:40:35+02:00

The race-car's motion round the track is a repeating motion that can be

described by a sinusoidal function.

The correct responses are;

Distance travelled when the car swept an angle of 212° ≈ 7.03 miles

If the car has travelled 4.9 miles, the angle swept out, θ ≈ 2.58 radians

The function is; [tex]\underline{h(t) = 1.9\cdot sin(45\cdot t)}[/tex].

Reasons:

Known parameters are;

Radius of the track, r = 1.9 miles

Point the car starts = 3 O'clock

Speed of the car = 85.5 mph

Required:

Distance swept out when the car traveled an angle of 212°.

Solution;

Distance of one complete turn = 2·π·r

Angle of one complete turn = 360°

Therefore, at 212°, we have;

[tex]\dfrac{212^{\circ}}{360^{\circ}} \times 2 \times 1.9 \times \pi \approx 7.03[/tex]

Distance travelled when the car swept an angle of 212° ≈ 7.03 miles

Required:

The measure of the angle swept out by the car if the car has travelled 4.9 miles.

Solution;

Let, θ represent the angle, we have;

[tex]\dfrac{\theta}{2 \cdot \pi} \times 2 \times 1.9 \times \pi \approx 4.9 \ miles[/tex]

We get;

[tex]{\theta} = \dfrac{4.9 \ miles}{2 \times 1.9 \ miles \times \pi } \times 2\cdot \pi \approx 2.58 \ radians[/tex]

If the car has travelled 4.9 miles, the angle swept out, θ ≈ 2.58 radians

Required:

The function, h, that gives the distance of the above the horizontal

diameter of the track (in miles) in terms of the number of hours since the

race-car started driving, t.

Solution;

The function that gives the car height can be presented as follows;

The angular velocity, ω = [tex]\dfrac{v}{r}[/tex]

Therefore;

[tex]\omega = \dfrac{85.5 \ mph}{1.9 \ miles} = 45 \ rad/hour[/tex]

The general form of the sinusoidal function is h = A·sin[k·(θ - b)] + c

Therefore, we get;

h = 1.9·sin[k·(θ - b)] + c

The period, T = 2·π/ω = 2·π/k

Therefore, given that ω·t = θ, we get;

h = 1.9·sin[ω·t - b] + c = 1.9·sin[45·t - b] + c

The vertical sift, c, and the horizontal shift, b, can both be taken as zero,

given that the sin(0) = 0, therefore;

h = 1.9·sin[0] + c = c = 0, such that the at the start, t = 0, the car is at a

distance of h = 0 above the horizontal line.

The function, h, that gives the distance of the car above the

horizontal track, in terms of the number of hours since the car started

driving, t, is therefore;

[tex]\underline{h(t) = 1.9\cdot sin(45\cdot t)}[/tex].

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