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?
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 start by filling out what you know and seeing what you need to solve for, the variables you need to be concerned with are:
$\displaystyle x_0 = \text{initial position}$
$\displaystyle x_f = \text{final position}$
$\displaystyle v_0 = \text{initial velocity}$
$\displaystyle v_f = \text{final velocity}$
$\displaystyle a = \text{acceleration}$
$\displaystyle t = \text{time}$
We know that:
$\displaystyle x_0 = 0 ~m$
$\displaystyle x_f = 40.0 ~m$
$\displaystyle v_0 = ? $
$\displaystyle v_f = 3.30 ~\frac{m}{s}$
$\displaystyle a = ?$
$\displaystyle t = 9.00 ~s$
For part a, you're going to want to use this equation:
$\displaystyle x_f = \frac{v_0 + v_f}{2}*t$
Just solve for $\displaystyle v_0$.
You can only solve part b after you have solved part a, and you will use this equation:
$\displaystyle v_f = v_0 + at$
Using the $\displaystyle v_0$ you got from part a. And you will be done.
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:
$\displaystyle x = ut + \frac{1}{2} a t^2$.
$\displaystyle x = vt - \frac{1}{2} a t^2$.
$\displaystyle 2ax = v^2 - u^2$.
$\displaystyle v = u + at$.
$\displaystyle x = \frac{(u + v) t}{2}$.
$\displaystyle {\text{Displacement }}\Delta \overrightarrow d = 40{\text{ m}} \hfill \\$
$\displaystyle {\text{Initial velocity }}\overrightarrow {v_1 } = ? \hfill \\$
$\displaystyle {\text{Final velocity }}\overrightarrow {v_2 } = 3.30{\text{ m/s}} \hfill \\$
$\displaystyle {\text{Time interval }}\Delta t = 9{\text{ s}}{\text{.}} \hfill \\$
$\displaystyle \left( a \right){\text{ Displacement, }}\Delta \overrightarrow d = \left( {\frac{{\overrightarrow {v_1 } + \overrightarrow {v_2 } }}
{2}} \right)\Delta t \hfill \\$
$\displaystyle 40 = \frac{{\overrightarrow {v_1 } + 3.30}}
{2} \times 9 \hfill \\$
$\displaystyle 40 \times \frac{2}
{9} = \overrightarrow {v_1 } + 3.30 \hfill \\$
$\displaystyle \overrightarrow {v_1 } = 5.59 \hfill \\$
$\displaystyle {\text{Initial velocity }}\overrightarrow {v_1 } = 5.59{\text{ m/s}}{\text{.}} \hfill \\$
$\displaystyle \left( b \right){\text{ Now, Acceleration, }}\overrightarrow a = \frac{{\overrightarrow {v_2 } - \overrightarrow {v_1 } }}
{{\Delta t}} \hfill \\$
$\displaystyle \overrightarrow a = \frac{{5.59 - 3.30}}
{9} = 0.26{\text{ m/s}}^2 \hfill \\$
$\displaystyle {\text{Acceleration, }}\overrightarrow a = 0.26{\text{ m/s}}^2 \hfill \\
$