Answer :
To stop a car going 75 mph on wet concrete with a deceleration of 5.00 m/s^2, it will travel approximately 112.09 meters before coming to a halt.
To determine the distance it takes for a car traveling at 75 mph to come to a full stop on wet concrete with an acceleration of -5.00 m/s2, we must first convert the speed to meters per second (m/s). 1 mph is equivalent to 0.44704 m/s, so 75 mph is 75 × 0.44704 m/s, which equals 33.528 m/s.
Next, we use the equation of motion d = vit + frac{1}{2}at2, where vi is the initial velocity, a is the acceleration, and d is the distance. Since the car is coming to a stop, the final velocity (vf) will be 0 m/s. We can use the equation vf2 = vi2 + 2ad to find the distance instead, where a is -5.00 m/s2 (deceleration is considered a negative acceleration).
Plugging in the values we get: 0 = (33.528 m/s)2 + 2 × (-5.00 m/s2) × d.
Solving for d gives us d = frac{(33.528 m/s)2}{2 × 5.00 m/s2} = 112.09 meters.
Therefore, the car will travel approximately 112.09 meters before coming to a halt on wet concrete when starting at a speed of 75 mph.
