A speedboat moving at 31.0 m/s approaches a no-wake buoy marker 100 m ahead. the pilot slows the boat with a constant acceleration of −4.00 m/s2 by reducing the throttle. (a) how long does it take the boat to reach the buoy? s (b) what is the velocity of the boat when it reaches the buoy? m/s
u = 31 m/s, the initial speed of the boat s = 100 m, distance to the buoy a = -4 m/s², the acceleratin (actually deceleration)
Part (a) Let t = time required to reach the buoy. Use the formula ut + (1/2)at² = s. (31 m/s)*(t s) - (1/2)*(4 m/s²)*(t s)² = (100 m) 2t² - 31t + 100 = 0
Solve with the quadratic formula. t = (1/4) [31 +/- √(31² - 800)] = 10.92 s or 4.58 s Before selecting the answer, we should determine the velocity at the buoy.
Part (b) When t = 10.92 s, the velocity at the buoy is v = (31 m/s) - (4 m/s²)*(10.92 s) = -12.68 m/s Because of the negative value, this value of t should be rejected.
When t = 4.58 s, the velocity at the buoy is v = (31 m/s) - (4 m/s²)*(4.58 s) = 12.68 m/s This value of t is acceptable.
Answer (to nearest tenth): (a) The time is 4.6 s (b) The velocity is 12.7 m/s
