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

Acceleration is a fundamental concept in physics that measures how an object's velocity changes over time. In the context of the problem provided, the car's initial velocity (vi) is 16 meters per second, and it comes to a complete stop with a final velocity (vf) of 0 meters per second in 4 seconds. The acceleration of the car during this time period can be calculated using the formula:
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.
← A penny dropped from a 700 ft high bridge Density calculation of an unknown liquid →

More Posts: