Acceleration and Distance Traveled by a Car in Uniform Motion
What is the distance traveled by a car that is brought to rest uniformly in 4.0 seconds?
The car in a time of 4 seconds will travel a distance of 32 meters.
Understanding Acceleration and Distance Traveled
a = (vf - vi) / t Plugging in the values: a = (0 m/s - 16 m/s) / 4 s a = -16 m/s / 4 s a = -4 m/s² The negative sign indicates that the car is decelerating or slowing down uniformly at a rate of 4 meters per second squared. Now, to determine the distance (x) traveled by the car during this 4.0-second interval, we can use the formula:
x = (vf² - vi²) / (2a) Substitute the known values: x = [(0 m/s)² - (16 m/s)²] / (2 * -4 m/s²) x = [0 m²/s² - 256 m²/s²] / -8 m/s² x = -256 m²/s² / -8 m/s² x = 32 meters Therefore, the car travels a distance of 32 meters during the 4.0-second interval while decelerating uniformly. This calculation demonstrates the relationship between acceleration, velocity, and distance traveled in uniformly varied rectilinear motion.