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A bar is traveling at 21.0 m/s. It slows to a stop at a constant rate over 5.00 seconds. How far does the car travel during those 5.00 seconds before it stops?

Answer :

Final answer:

To find the distance a car travels while slowing down at a constant rate, we can use the formula distance = initial velocity * time + 0.5 * acceleration * time^2. By rearranging the formula and plugging in the values, we can calculate the distance traveled.


Explanation:

To find the distance the car travels during the 5 seconds, we can use the formula:

distance = initial velocity * time + 0.5 * acceleration * time^2

Since the car is coming to a stop, its final velocity is 0 m/s. We can assume that the acceleration is constant, so we can rearrange the formula:

distance = initial velocity * time + 0.5 * acceleration * time^2 = (initial velocity * time) - 0.5 * acceleration * time^2

Plugging in the values, we have:

distance = (21.0 m/s * 5.00 s) - 0.5 * acceleration * (5.00 s)^2

Since we need to solve for distance, we can set up the equation:

distance = (21.0 m/s * 5.00 s) - 0.5 * distance = (105.0 m) - 0.5 * acceleration * 25.0 s^2

We don't know the value of acceleration, but we can solve for it by rearranging the equation:

0.5 * acceleration * 25.0 s^2 = 105.0 m - distance

acceleration = (105.0 m - distance) / (0.5 * 25.0 s^2)

Now we can substitute this value back into the distance equation:

distance = (21.0 m/s * 5.00 s) - 0.5 * ((105.0 m - distance) / (0.5 * 25.0 s^2)) * (5.00 s)^2

Solving this equation will give us the distance the car travels during the 5.00 seconds before coming to a stop.


Learn more about distance traveled while slowing down here:

https://brainly.com/question/27996914


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