1. ## Linear Motion Problem

A truck covers 40.0 m in 9.00 s while smoothly slowing down to a final speed of 3.30 m/s.

A) Find the truck's original speed.

B) Find the truck's acceleration.

I don't know where to start, can anyone point me in the right direction?

2. You start by filling out what you know and seeing what you need to solve for, the variables you need to be concerned with are:

$x_0 = \text{initial position}$
$x_f = \text{final position}$
$v_0 = \text{initial velocity}$
$v_f = \text{final velocity}$
$a = \text{acceleration}$
$t = \text{time}$

We know that:

$x_0 = 0 ~m$
$x_f = 40.0 ~m$
$v_0 = ?$
$v_f = 3.30 ~\frac{m}{s}$
$a = ?$
$t = 9.00 ~s$

For part a, you're going to want to use this equation:

$x_f = \frac{v_0 + v_f}{2}*t$

Just solve for $v_0$.

You can only solve part b after you have solved part a, and you will use this equation:

$v_f = v_0 + at$

Using the $v_0$ you got from part a. And you will be done.

3. Originally Posted by Pn0yS0ld13r
A truck covers 40.0 m in 9.00 s while smoothly slowing down to a final speed of 3.30 m/s.

A) Find the truck's original speed.

B) Find the truck's acceleration.

I don't know where to start, can anyone point me in the right direction?
You can start by recognising that you need to use the formulae for uniform straight line motion. Then you can look at the given data:

x = 40.0 m

t = 9.00 s

v = 3.30 m/s

A) u = ?

B) a = ?

In each case, use a formula that links the given data to the unknown quantity. Choose from the following:

$x = ut + \frac{1}{2} a t^2$.

$x = vt - \frac{1}{2} a t^2$.

$2ax = v^2 - u^2$.

$v = u + at$.

$x = \frac{(u + v) t}{2}$.

4. Originally Posted by Pn0yS0ld13r
A truck covers 40.0 m in 9.00 s while smoothly slowing down to a final speed of 3.30 m/s.

A) Find the truck's original speed.

B) Find the truck's acceleration.

I don't know where to start, can anyone point me in the right direction?
${\text{Displacement }}\Delta \overrightarrow d = 40{\text{ m}} \hfill \\$

${\text{Initial velocity }}\overrightarrow {v_1 } = ? \hfill \\$

${\text{Final velocity }}\overrightarrow {v_2 } = 3.30{\text{ m/s}} \hfill \\$

${\text{Time interval }}\Delta t = 9{\text{ s}}{\text{.}} \hfill \\$

$\left( a \right){\text{ Displacement, }}\Delta \overrightarrow d = \left( {\frac{{\overrightarrow {v_1 } + \overrightarrow {v_2 } }}
{2}} \right)\Delta t \hfill \\$

$40 = \frac{{\overrightarrow {v_1 } + 3.30}}
{2} \times 9 \hfill \\$

$40 \times \frac{2}
{9} = \overrightarrow {v_1 } + 3.30 \hfill \\$

$\overrightarrow {v_1 } = 5.59 \hfill \\$

${\text{Initial velocity }}\overrightarrow {v_1 } = 5.59{\text{ m/s}}{\text{.}} \hfill \\$

$\left( b \right){\text{ Now, Acceleration, }}\overrightarrow a = \frac{{\overrightarrow {v_2 } - \overrightarrow {v_1 } }}
{{\Delta t}} \hfill \\$

$\overrightarrow a = \frac{{5.59 - 3.30}}
{9} = 0.26{\text{ m/s}}^2 \hfill \\$

${\text{Acceleration, }}\overrightarrow a = 0.26{\text{ m/s}}^2 \hfill \\
$

5. Thank you guys; I got it now.

6. Originally Posted by Shyam
$\left( a \right){\text{ Displacement, }}\Delta \overrightarrow d = \left( {\frac{{\overrightarrow {v_1 } + \overrightarrow {v_2 } }}
{2}} \right)\Delta t \hfill \\$
Just to prove I'm picky, the quantity $\Delta d$ is a vector, so the symbol should be $\overrightarrow{ \Delta d}$.

-Dan