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Correct answers only please! If you don't know the answer, then please don't guess or say what you think it is.

Mike recently increased the size of his truck tires from the original P215/60R16 to the larger P235/7016. If Mike didn't recalibrate his speedometer, how fast is really going on the new tires when his speedometer shows he is traveling 60 mph?

A. 54.2 mph

B. 63.8 mph

C. 66.4 mph

D. 69.7 mph

Correct answers only please If you dont know the answer then please dont guess or say what you think it is Mike recently increased the size of his truck tires f class=

Respuesta :

Answer:

A. 54.2 mph

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

P215/60R16 meaning

215 is width in millimeters

60 is the ratio of the height of the tire’s cross-section to its width

16 is wheel diameter

So rim height is

60*215/100 = 129 mm = 5 inches

Then, total diameter of tyre is

16'' + 5'' = 21''

that means radius is 21''/2 = 10.5''

The new tyre P235/70R16 have a height of

70*235/100 = 164.5 mm = 6.5''

and its total diameter is

16'' + 6.5'' = 22.5''

that means radius is 22.5''/2 = 11.25''

We know that linear velocity (v) is the radius (r) times the angular velocity ([tex]\omega[/tex]). For the first case

[tex] v_1 = r_1 \times \omega_1[/tex]

And for the second one

[tex] v_2 = r_2 \times \omega_2[/tex]

Angular velocity doesn't change from the first case to the second case, i. e., [tex]\omega_1 = \omega_2[/tex]. Combining the equations and replacing with values give

[tex] \frac{v_1}{r_1} = \frac{v_2}{r_2}[/tex]

[tex] \frac{v_1}{r_1} \times r_2 = v_2[/tex]

[tex] \frac{60 mph}{10.5} \times 11.25 = v_2[/tex]

[tex] v_2 = 63.8 mph[/tex]