Question

Question: An iceboat has a constant velocity toward the east when a sudden gust of wind causes the iceboat to have a constant acceleration toward the east for a period of$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$. A plot of x versus t is shown in Fig. 2-47, where $\mathit{t}\mathbf{=}\mathbf{0}$is taken to be the instant the wind starts to blow and the positiveis toward the east. (a) What is the acceleration of the iceboat during the$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$interval? (b) What is the velocity of the iceboat at the end of the$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$interval? (c) If the acceleration remains constant for an additional$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$, how far does the iceboat travel during this second$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$interval?

Short answer

1. The acceleration is $2.0\mathrm{m}/{\mathrm{s}}^2$.
2. The velocity of the iceboat at the end of the $3.0\mathrm{s}$interval is$12\mathrm{m}/\mathrm{s}.$
3. The iceboat travels$45\mathrm{m}$during the second interval when the acceleration is constant.

Step-by-step solution

Given information

$\begin{array}{l}{\mathrm{x}}_1=16\mathrm{m}\\ {\mathrm{x}}_2=27\mathrm{m}\\ \mathrm{t}=2\mathrm{s}\\ {\mathrm{t}}_2=3\mathrm{s}\end{array}$

To understand the concept

The problem deals with the kinematic equation of motion. Kinematics is the study of how a system of bodies moves without taking into account the forces or potential fields that influence the motion. The equations which are used in the study are known as kinematic equations of motion.

Formula:

The displacement in the kinematic equation is given by,

$\mathrm{x}={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2$

(a): Calculation for the acceleration

Using equation (i), the displacement can be written as

$$\begin{gathered} {\mathrm{x}}_1={\mathrm{v}}_0\times {\mathrm{t}}_1+\frac{1}{2}\times \mathrm{a}\times {\mathrm{t}}_1^2 \\ {\mathrm{x}}_2={\mathrm{v}}_0\times {\mathrm{t}}_2+\frac{1}{2}\times \mathrm{a}\times {\mathrm{t}}_2^2 \\ 16={\mathrm{v}}_{\mathrm{i}}\times 2.0+\frac{1}{2}\times \mathrm{a}\times 2.0^2 \\ 27={\mathrm{v}}_{\mathrm{i}}\times 3.0+\frac{1}{2}\times \mathrm{a}\times 3.0^2 \end{gathered}$$

After solving the above two equations,

${\mathrm{v}}_0=6.0\mathrm{m}/\mathrm{s}$

$\mathrm{a}=2.0\mathrm{m}/{\mathrm{s}}^2$

So, the acceleration would be$2.0\mathrm{m}/{\mathrm{s}}^2$.

(b): Calculation for the velocity of iceboat at the end of the interval

The speed at end of $3.0\mathrm{s}$interval is given by

$$\begin{gathered} {\mathrm{x}}_2={\mathrm{v}}_0{\mathrm{t}}_2-\frac{1}{2}{{\mathrm{at}}_2}^2 \\ \Rightarrow 27={\mathrm{v}}_0\times 3.0-\frac{1}{2}\times \mathrm{a}\times 3.0^2 \\ \Rightarrow {\mathrm{v}}_0=12\mathrm{m}/\mathrm{s} \end{gathered}$$

So, the velocity of iceboat at the end of the $3.0\mathrm{s}$interval is$12\mathrm{m}/\mathrm{s}$.

(c): Calculations for distance traveled by iceboat during

Now, the distance it traveled in $3.0\mathrm{s}$

$$\begin{gathered} \mathrm{x}={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{x}=12\times 3+\frac{1}{2}\times 2\times 3^2 \\ \mathrm{x}=45\mathrm{m} \end{gathered}$$

So, the distance traveled within $3.0\mathrm{s}$is $45\mathrm{m}$.