Question

A motorboat accelerates uniformly from a velocity of 6.5 m/s west to a velocity of 1.5 m/s west. If its acceleration was 2.7 m/s2 to the east, what is the displacement
of the motorboat during the acceleration?

Answer 1

Answer:

The motorboat ends up 7.41 meters to the west of the initial position

Explanation:

Accelerated Motion

The accelerated motion describes a situation where an object changes its velocity over time. If the acceleration is constant, then these formulas apply:

$\vec v_f=\vec v_o+\vec a.t$

$\displaystyle \vec r=\vec v_o.t+\frac{\vec a.t^2}{2}$

The problem provides the conditions of the motorboat's motion. The initial velocity is 6.5 m/s west. The final velocity is 1.5 m/s west, and the acceleration is $2.7 m/s^2$ to the east. Since all the movement takes place in one dimension, we can ignore the vectorial notation and work with the signs of the variables, according to a defined positive direction. We'll follow the rule that all the directional magnitudes are positive to the east and negative to the west. Rewriting the formulas:

$v_f=v_o+a.t$

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Solving the first one for t

$\displaystyle t=\frac{v_f-v_o}{a}$

We have

$v_o=-6.5,\ v_f=-1.5,\ a=2.7$

Using these values

$\displaystyle t=\frac{-1.5+6.5}{2.7}=1.852\ s$

We now compute x

$\displaystyle x=(-6.5)(1.852)+\frac{(2.7)(1.852)^2}{2}$

$x=-7,41\ m$

The motorboat ends up 7.41 meters to the west of the initial position