The driver of a car traveling at 20.7m/s applies the brakes and undergoes a constant deceleration of 2.85m/s^2 How many revolutions does each tire make before the car comes to a stop.assuming that the car does not skid and that the tires have radii of 0.26 m? Answer in units of rev.Answer in units of rev.

To solve this problem, we need to find the distance the car travels before coming to a stop, and then calculate the number of revolutions each tire makes during that distance.<br /><br />Given information:<br />- Initial velocity of the car: $v_0 = 20.7 \, \text{m/s}$<br />- Deceleration of the car: $a = -2.85 \, \text{m/s}^2$ (negative because the car is slowing down)<br />- Radius of the tires: $r = 0.26 \, \text{m}$<br /><br />Step 1: Calculate the distance the car travels before coming to a stop.<br />We can use the kinematic equation:<br />$v^2 = v_0^2 + 2a s$<br />where $v$ is the final velocity, $v_0$ is the initial velocity, $a$ is the acceleration, and $s$ is the distance traveled.<br /><br />Since the car comes to a stop, the final velocity $v$ is 0.<br />$0 = (20.7 \, \text{m/s})^2 + 2(-2.85 \, \text{m/s}^2) s$<br />$s = \frac{(20.7 \, \text{m/s})^2}{2(2.85 \, \text{m/s}^2)}$<br />$s \approx 146.6 \, \text{m}$<br /><br />Step 2: Calculate the number of revolutions each tire makes.<br />The circumference of a circle is given by $C = 2\pi r$, where $r$ is the radius.<br />$C = 2\pi \times 0.26 \, \text{m}$<br />$C \approx 1.63 \, \text{m}$<br /><br />The number of revolutions each tire makes is the total distance traveled divided by the circumference of the tire.<br />$\text{Number of revolutions} = \frac{\text{Distance traveled}}{\text{Circumference of the tire}}$<br />$\text{Number of revolutions} = \frac{146.6 \, \text{m}}{1.63 \, \text{m}}$<br />$\text{Number of revolutions} \approx 90.2$<br /><br />Therefore, each tire makes approximately 90.2 revolutions before the car comes to a stop.