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A car moves at a constant velocity of 24(m)/(s), brakes so that it has a constant deceleration of 0.952 s^(2). Determine the speed of the car after a distance of 250 meters.

User Newtopian
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Final answer:

Using the physics equation for kinetics, the final velocity of the car, given a distance of 250 meters and a deceleration of 0.952 m/s^2 from an initial velocity of 24 m/s is found to be 10 m/s.

Step-by-step explanation:

The question is about a car moving at a constant velocity and then breaks to have a constant deceleration. We need to find the velocity of the car after it has traveled a certain distance while decelerating. We use the equation v^2 = u^2 - 2as, where v is the final velocity, u is the initial velocity, a is the acceleration (in this case, deceleration), and s is the distance traveled.

In this case, u = 24 m/s (the initial velocity), a = -0.952 m/s^2 (the deceleration, negative because it is slowing down), and s = 250 m (the distance traveled). Substituting these values in, we get v^2 = (24 m/s)^2 - 2 * -0.952 m/s^2 * 250 m = 576 m^2/s^2 - 476 m^2/s^2 = 100 m^2/s^2. Taking the square root of 100 m^2/s^2 gives us the final velocity v = 10 m/s.

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User Matt Jacobsen
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