A speedboat moving at 40.0 m/s approaches a no-wake buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of -3.30 m/s^2 by reducing the throttle. (a) How long does it take the boat to reach the buoy? (b) What is the velocity of the boat when it reaches the buoy?
Added by Meredith C.
Step 1
0\) m/s \(a = -3.30\) m/s\(^2\) Substitute the values into the equation: \(100 = 40.0t + \frac{1}{2}(-3.30)t^2\) Solve for \(t\): \(-1.65t^2 + 40t - 100 = 0\) Using the quadratic formula, we get: \(t = 2.83\) seconds ** Show more…
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