What is the maximum speed (in meters/second) with which a 1300 kg car can round a turn of radius 700 m on a flat road, if the coefficent of friction between tires and road is 0.530
To determine the maximum speed with which the car can round the turn, we need to consider the centripetal force acting on the car and the maximum frictional force that can be exerted by the tires.
The centripetal force required to keep the car moving in a circular path is given by the equation:
Fc = (mass × velocity^2) / radius
where Fc is the centripetal force, mass is the mass of the car, velocity is the speed of the car, and radius is the radius of the turn.
The maximum frictional force that can be exerted by the tires is given by the equation:
Ff = coefficient of friction × normal force
where Ff is the frictional force, coefficient of friction is the given value of 0.530, and normal force is the force exerted by the road on the tires.
In this case, the normal force is equal to the weight of the car, which is given by:
Normal force = mass × gravitational acceleration
where gravitational acceleration is approximately 9.8 m/s^2.
Now, we can set up an inequality to find the maximum speed:
Ff ≥ Fc
(coefficent of friction × normal force) ≥ (mass × velocity^2) / radius
(0.53 × (mass × gravitational acceleration)) ≥ (mass × velocity^2) / radius
Simplifying further:
0.53 × (mass × gravitational acceleration) ≥ (mass × velocity^2) / radius
0.53 × mass × gravitational acceleration ≥ mass × velocity^2 / radius
0.53 × gravitational acceleration ≥ velocity^2 / radius
To find the maximum velocity, we can rearrange the equation:
velocity^2 ≤ 0.53 × gravitational acceleration × radius
velocity ≤ sqrt(0.53 × gravitational acceleration × radius)
Plugging in the given values:
velocity ≤ sqrt(0.53 × 9.8 m/s^2 × 700 m)
velocity ≤ sqrt(0.53 × 9.8 × 700) m/s
velocity ≤ sqrt(3519.4) m/s
velocity ≤ 59.3 m/s
Therefore, the maximum speed with which a 1300 kg car can round the turn is approximately 59.3 meters/second.