Answer to Question #135411 in Mechanics | Relativity for Rashinda

Question

A speedboat moving at 30.0 m/s approaches a ship 100 m ahead. The pilot slows the boat with a constant acceleration of -3.50 m/s2. How long does it take the boat to reach the ship?

Answer 1

When the boat is moving with constant acceleration, its position depends on time as

$\displaystyle s = v_0t + \frac{at^2}{2}$

Let's put into it $s=100\, m$, $v_0=30.0 \, m/s$ , $a = -3.50 \,m/s^2$.

$100 = 30t -1.75 t^2$

$1.75 t^2 - 30t + 100=0$

$\displaystyle t_{1,2} = \frac{30 \pm \sqrt{900-700}}{3.5}= \frac{30\pm14.14}{3.5}=4.53 \;s$

We have chosen minus sign here because of the properties of parabola. s(t) is parabola. We start at some point, move to the apex of a parabola and then return back. We need to chose the first time when t satisfies the equation, namely, the smaller root.

Answer: $t = 4.53\,s$