Question
A speedboat moving at 29 m/s approaches a no-wake buoy marker 98.3 m ahead. The pilot slows the boat with a constant acceleration of -3.61 m/s2 by reducing the throttle. How long does it take the boat to reach the buoy? (in s)
Answers
Damon
d = Vi t + (1/2) a t^2
98.3 = 29 t - 1.8 t^2
1.8 t^2 -29 t + 98.3 = 0
solve quadratic for t
[ 29 +/- sqrt ( 841 - 708) ] /3.6
[ 29 +/- 11.5 ] / 3.6
= 4.86 (ignore the big time, it is on the way back :)
98.3 = 29 t - 1.8 t^2
1.8 t^2 -29 t + 98.3 = 0
solve quadratic for t
[ 29 +/- sqrt ( 841 - 708) ] /3.6
[ 29 +/- 11.5 ] / 3.6
= 4.86 (ignore the big time, it is on the way back :)