Answer to Question #165990 in Physics for izha

Question

 A speedboat moving at 30.0 m/s approaches a no-wake buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of 23.50 m/s2 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? 

Answer 1

(a) We can find the time that the boat takes to reach the buoy from the kinematic equation (it seems there is a typo in the boat's acceleration value, because for that value the roots of quadratic equation don't exist):

$$s=v_0t+\dfrac{1}{2}at^2,$$ $$100=30t+0.5\cdot(-3.50\ \dfrac{m}{s^2})t^2,$$ $$1.75t^2-30t+100=0.$$

This quadratic equation has two roots: $t_1=4.53\ s$ and $t_2=12.61\ s$. The correct answer is $t=4.53\ s.$

(b) We can find the velocity of the boat when it reaches the buoy from another kinematic equation:

$$v=v_0+at,$$ $$v=30\ \dfrac{m}{s}+(-3.5\ \dfrac{m}{s^2})\cdot4.53\ s=14.14\ \dfrac{m}{s}.$$