Question

a motorboat accelerates uniformly from a velocity of 6.5 m/s to the west to a velocity of 1.5 m/s to the west. if its acceleration was 2.7 m/s² to the east, how far did it travel during the acceleration?

Explanation:

Step1: Identify the kinematic - equation

We use the equation $v^{2}=v_{0}^{2}+2as$. Here, $v$ is the final velocity, $v_{0}$ is the initial velocity, $a$ is the acceleration and $s$ is the displacement.

Step2: Define the variables

The initial velocity $v_{0}=6.5\ m/s$ (east, so we can take it as positive), the final velocity $v = - 1.5\ m/s$ (west, negative), and the acceleration $a=-2.7\ m/s^{2}$ (negative as it is in the west - direction).

Step3: Rearrange the kinematic - equation for displacement

From $v^{2}=v_{0}^{2}+2as$, we can solve for $s$: $s=\frac{v^{2}-v_{0}^{2}}{2a}$.

Step4: Substitute the values

\[

$$\begin{align*} s&=\frac{(-1.5)^{2}-(6.5)^{2}}{2\times(-2.7)}\\ &=\frac{2.25 - 42.25}{- 5.4}\\ &=\frac{-40}{-5.4}\\ &\approx7.41\ m \end{align*}$$

\]
To find the acceleration, we can also use the equation $v = v_{0}+at$. First, we find the time $t=\frac{v - v_{0}}{a}=\frac{-1.5 - 6.5}{-2.7}=\frac{-8}{-2.7}\approx2.96\ s$. And we already know the acceleration $a=-2.7\ m/s^{2}$.

Answer:

The distance traveled $s\approx7.41\ m$ and the acceleration $a = - 2.7\ m/s^{2}$