MHB How far will the car travel in 10 seconds

[karush]

$\tiny{Embry-Brittle \, 12}$
$\textsf{A car, starting from rest, accelerates in a straight line at a constant rate of $\displaystyle 2.0 \frac{m}{s^2}$}$$\textit{How far will the car travel in $10$ seconds}$\begin{align*}\displaystyle
\Delta t&=10\\
a&=2\\
d_i&=0\\
v_i&=0\\
d_f&= d_i + v_i\Delta t + \frac{1}{2} a \Delta t^2\\
&=0+0\cdot 10+\frac{1}{2} \cdot 2\cdot 10^2\\
&=\color{red}{100 \, m}
\end{align*}no answer given so hope this is ok

 

[Prove It]

$\tiny{Embry-Brittle \, 12}$
$\textsf{A car, starting from rest, accelerates in a straight line at a constant rate of $\displaystyle 2.0 \frac{m}{s^2}$}$$\textit{How far will the car travel in $10$ seconds}$\begin{align*}\displaystyle
\Delta t&=10\\
a&=2\\
d_i&=0\\
v_i&=0\\
d_f&= d_i + v_i\Delta t + \frac{1}{2} a \Delta t^2\\
&=0+0\cdot 10+\frac{1}{2} \cdot 2\cdot 10^2\\
&=\color{red}{100 \, m}
\end{align*}no answer given so hope this is ok

$\displaystyle \begin{align*} v &= \int{ a\,\mathrm{d}t} \\ v &= a\,t + C_1 \\ v &= a\,t \textrm{ since the car starts from rest...} \\ x &= \int{ v\,\mathrm{d}t} \\ x &= \int{ a\,t \, \mathrm{d}t } \\ x &= \frac{1}{2}\,a\,t^2 + C_2 \textrm{ where } C_2 \textrm{ is the starting position...} \end{align*}$

So with $\displaystyle \begin{align*} t = 10 \textrm{ and } a = 2 \end{align*}$ we have

$\displaystyle \begin{align*} x &= \frac{1}{2} \cdot 2 \cdot 10^2 + C_2 \\ &= 100 + C_2 \end{align*}$

Thus the car has traveled 100 metres.

 

[karush]

ok
I didn't know you could us a integral on itwhy did you shift from $v=$ to $x=$

 

[MarkFL]

You could use the kinematic equation:

$$x=\frac{v_f^2-v_i^2}{2a}=\frac{\left(20\frac{\text{m}}{\text{s}}\right)^2-\left(0\frac{\text{m}}{\text{s}}\right)^2}{2\left(2\frac{\text{m}}{\text{s}^2}\right)}=100\text{ m}$$

 

[Greg]

why did you shift from $v=$ to $x=$

$v$ is velocity, $x$ is position.