Answer to Question #225887 in Physics for aimi

Question

a speedboat moving 30ms-1 approach stationary buoy marker 100m ahead. the pilot slows the boat with a constant acceleration of -3.50ms-2 by reducing the throttle. how long does it take the boat to reach the buoy?

Answer 1

The distance covered by the boat is given by the kinematic equation:

$$d = v_0t -\dfrac{at^2}{2}$$

where $d = 100m$, $v_0 = 30 m/s$ and $a = 3.5m/s^2$. Solving the quadratic equation for $t$, obtain:

$$1.75t^2 - 30t + 100 = 0\\ t \approx 4.5s$$

Here only first solution was taken, since this is the closest time the boat reach the buoy. The second time will be on its way back, if it maintain the same acceleration.

Answer. 4.5 s.