The wheel makes approximately [tex]\(\boxed{37.14}\)[/tex] revolutions before the car comes to a complete stop. This result is derived by first calculating the total distance traveled during deceleration and then determining the number of revolutions based on the wheel's circumference.
The problem involves determining the number of revolutions a car's wheel makes before it comes to a complete stop. The car initially travels at 30 m/s and stops over a period of 7 seconds. The wheel radius is given as 45 cm. This solution will employ kinematic equations to find the total distance traveled by the car and then use that distance to calculate the number of wheel revolutions.
To find the distance, we'll use the kinematic equations of motion. The initial velocity [tex]\( u \)[/tex] is 30 m/s, the final velocity [tex]\( v \)[/tex] is 0 m/s (as the car stops), and the time taken [tex]\( t \)[/tex] is 7 seconds. The acceleration [tex]\( a \)[/tex] can be found using the equation:
[tex]$a = \frac{v - u}{t}[/tex]
Substituting the values:
[tex]$a = \frac{0 - 30}{7} = -4.29\, \text{m/s}^2[/tex]
Now, using the equation [tex]\( s = ut + \frac{1}{2}at^2 \)[/tex] where [tex]\( s \)[/tex] is the distance, we find:
[tex]$\begin{align*}s &= 30 \times 7 + \frac{1}{2} \times (-4.29) \times 7^2\\[1em]&= 105\, \text{m}\end{align*}[/tex]
The circumference of the wheel is given by [tex]\( C = 2\pi r \)[/tex], where [tex]\( r \)[/tex] is the radius of the wheel (0.45 m). The number of revolutions [tex]\( N \)[/tex] is the total distance divided by the circumference:
[tex]$\begin{align*}C &= 2\pi \times 0.45\\[1em]N &= \frac{s}{C} = \frac{105}{2\pi \times 0.45}\end{align*}[/tex]
After performing the calculations, we find:
[tex]$N = \boxed{37.14 \text{ revolutions}}[/tex]
