.
To solve this problem, we can use the equation for centripetal acceleration, a = v^2/r, where a is the acceleration, v is the tangential velocity, and r is the radius of the circle. We know that the tangential acceleration is given as 1.7m/s^2, and we can assume that the car was traveling at a constant speed around the circle before it skidded off. This means that the tangential velocity is also constant.
Since the car skids off the track after traveling one quarter of the way around, we can say that the radius of the circle is also one quarter of the total circumference. Therefore, we can set up the following equation:
1.7m/s^2 = (v^2)/(r/4)
Solving for v, we get v = 2.93m/s. Now, we can use the equation for centripetal force, Fc = mv^2/r, where m is the mass of the car. We know that the only force acting on the car is the force of friction between the car and the track, since there is no external force causing the car to accelerate tangentially. Therefore, we can set up the following equation:
Fc = μN = mv^2/r
Where μ is the coefficient of static friction and N is the normal force between the car and the track. We can rearrange this equation to solve for μ:
μ = (mv^2)/(rN)
To find N, we can use Newton's second law, F = ma, where F is the force of friction and a is the tangential acceleration. We know that the force of friction is equal to the force of gravity acting on the car, since the car is not moving vertically. Therefore, we can set up the following equation:
F = mg = ma
Solving for N, we get N = mg/a.
Substituting this into our previous equation, we get:
μ = (mv^2)/(rmg/a)
Simplifying, we get:
μ = (v^2)/(rg)
Plugging in the values we have calculated, we get:
μ = (2.93m/s)^2 / (1/4 * 9.8m/s^2 * 1/4 * πr)
μ = 0.54
Therefore, the coefficient of static friction between the car and the track is 0.54
