Question

homework 5 begin date: 8/17/2025 11:00:00 pm due date: 9/30/2025 11:59:00 pm end date: 9/30/2025 11:59:00 pm problem 9: (8% of assignment value) a car is driving along a level and unbanked circular track of diameter d at a constant speed of v. part (a) write an expression for the magnitude of the acceleration of the car in terms of the given parameters. a = 2 v^2/d correct! part (b) the diameter of the track and speed of the car are measured and it is found that d = 940 m and v = 29.7 m/s. what is the magnitude, in meters per squared second, of the acceleration of the car? a = | m/s^2 grade summary deductions 0% potential 100% submissions attempts remaining: 3 4% deduction per attempt detailed view

Explanation:

Step1: Identify the formula

The centripetal - acceleration formula for an object moving in a circular path is $a=\frac{v^{2}}{r}$, where $v$ is the speed and $r$ is the radius. Given the diameter $d$, we know that $r = \frac{d}{2}$. So the formula becomes $a=\frac{2v^{2}}{d}$.

Step2: Substitute the given values

We are given $d = 940$ m and $v=29.7$ m/s. Substitute these values into the formula $a=\frac{2v^{2}}{d}$. First, calculate $v^{2}=(29.7)^{2}=882.09$. Then $2v^{2}=2\times882.09 = 1764.18$. Now, $a=\frac{1764.18}{940}$.

Step3: Calculate the acceleration

$a=\frac{1764.18}{940}\approx1.88$ m/s².

Answer:

$1.88$ m/s²