Answer :
A. The acceleration of the car during the 5 seconds is -10 mi/h/s.
B. The car will stop after traveling a distance of 220.83 ft (or 67.31 meters).
C. The average velocity of the car between the braking point and the stopping point is 25 mi/h.
D. The velocity of the car will be 25 mi/h at a distance of 183.33 ft (or 55.88 meters) from the braking point.
A. To calculate the acceleration of the car, we can use the formula: acceleration = (final velocity - initial velocity) / time. The initial velocity is 50 mi/h, the final velocity is 0 mi/h (since the car stops), and the time is 5 seconds. By plugging these values into the formula, we find that the acceleration of the car during the 5 seconds is -10 mi/h/s. The negative sign indicates that the car is decelerating.
B. The distance traveled by the car during the 5 seconds can be calculated using the formula: distance = (initial velocity * time) + (0.5 * acceleration * time^2). Substituting the given values, we find that the car will stop after traveling a distance of 220.83 ft (or 67.31 meters).
C. The average velocity of the car between the braking point and the stopping point can be calculated by dividing the total distance traveled by the time taken. Since the car comes to a stop, the average velocity will be half of the initial velocity. Therefore, the average velocity is 25 mi/h.
D. To find the distance at which the velocity of the car is 25 mi/h, we can use the formula: distance = (initial velocity^2 - final velocity^2) / (2 * acceleration). Substituting the given values, we find that the velocity of the car will be 25 mi/h at a distance of 183.33 ft (or 55.88 meters) from the braking point.
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