Question

A car accelerates from rest and reaches a speed of 25m\s in 7.6s. if the diameter of a tire is 0.680 m , find the revolutions the tire makes during the motion?

Answer 1

Answer on Question #45693, Physics, Mechanics | Kinematics | Dynamics

A car accelerates from rest and reaches a speed of $25\mathrm{m/s}$ in 7.6s. If the diameter of a tire is $0.680\mathrm{m}$, find the revolutions the tire makes during the motion?

Solution:

The kinematic equation that describes an object's motion is:

$$d = \frac {v _ {f} ^ {2} - v _ {i} ^ {2}}{2 a}$$

The symbol $d$ stands for the displacement of the object. The symbol $a$ stands for the acceleration of the object. And the symbol $v$ stands for the velocity of the object; a subscript of $i$ after the $v$ indicates that the velocity value is the initial velocity value and a subscript of $f$ indicates that the velocity value is the final velocity value.

The acceleration is

$$a = \frac {v _ {f} - v _ {i}}{t}$$

In our case:

$$v _ {i} = 0 \mathrm {m / s},$$ $$v _ {f} = 25 \mathrm {m / s},$$ $$t = 7.6 \mathrm {s},$$

Thus,

$$d = \frac {v _ {f} t}{2} = \frac {25 \cdot 7.6}{2} = 95 \mathrm {m}$$

The circumference of a tire is

$$C = \pi D = 3.1415 \cdot 0.680 = 2.1363 \mathrm {m}$$

The number of revolutions the tire makes during the motion is

$$N = \frac {d}{C} = \frac {95}{2.1363} = 44.47$$

Answer: $N = 44.5$

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