# acceleration/speed question

• September 8th 2008, 06:20 PM
2clients
acceleration/speed question
The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with an acceleration of -5.60 m/s^2
for 4.20 s, making straight skid marks 62.4 m long ending at the tree. With what speed does the car then strike the tree?

I thought I understood acceleration and speed until this problem!

Thanks!
• September 8th 2008, 06:28 PM
skeeter
$\Delta x = v_f t - \frac{1}{2}at^2$

solve for $v_f$
• September 8th 2008, 06:31 PM
2clients
Quote:

Originally Posted by skeeter
$\Delta x = v_f t - \frac{1}{2}at^2$

solve for $v_f$

Thanks! Though I would be very grateful if you could say how you got there. I'm not sure what the variables are. I would love a few sentences!!
• September 8th 2008, 06:40 PM
11rdc11
$\Delta x = \text{final position - intital position}$

$v_f = \text{final velocity}$

$t = \text{time}$

$a = \text{acceleration}$
• September 8th 2008, 06:45 PM
2clients
Quote:

Originally Posted by 11rdc11
$\Delta x = \text{final position - intital position}$

$v_f = \text{final velocity}$

$t = \text{time}$

$a = \text{acceleration}$

Thanks! Is there a more general formula from which this formula is derived? Where does the 1/2 come from? Is it the formula for average acceleration?
• September 8th 2008, 06:54 PM
11rdc11
I think from this postion function

$x(t) = \frac{at^2}{2} + vt + x$

which when you take the derivative is

$v(t) = at +v$

and then

$a(t) = a$
• September 8th 2008, 07:36 PM
Shyam
Quote:

Originally Posted by 2clients
The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with an acceleration of -5.60 m/s^2
for 4.20 s, making straight skid marks 62.4 m long ending at the tree. With what speed does the car then strike the tree?

I thought I understood acceleration and speed until this problem!

Thanks!

${\text{Now, }}\overrightarrow a = {\text{acceleration}} = - 5.6{\text{ m/s}}^2 \hfill \\$

$\overrightarrow {v_1 } = {\text{ initial velocity of car}} \hfill \\$

$\overrightarrow {v_2 } = {\text{ final velocity of car}} \hfill \\$

$\Delta t = {\text{ time interval }} = 4.2{\text{ s}} \hfill \\$

$\Delta \overrightarrow d = {\text{ displacement of car }} = 62.4{\text{ m}}{\text{.}} \hfill \\$

${\text{Now, using the formula, Acceleration, }}\overrightarrow a = \frac{{\overrightarrow {v_2 } - \overrightarrow {v_1 } }}
{{\Delta t}} \hfill \\$

$\overrightarrow {v_2 } - \overrightarrow {v_1 } = - 5.6 \times 4.20 \hfill \\$

$\overrightarrow {v_2 } = \overrightarrow {v_1 } - 23.52{\text{ }}......................{\text{eqn}}{\text{. (1)}} \hfill \\$

${\text{Now, Displacement, }}\Delta \overrightarrow d = \overrightarrow {v_1 } \Delta t + \frac{1}
{2}\overrightarrow a \left( {\Delta t} \right)^2 \hfill \\$

$62.4 = \overrightarrow {v_1 } \left( {4.2} \right) + \frac{1}
{2}\left( { - 5.6} \right)\left( {4.2} \right)^2 \hfill \\$

$\overrightarrow {v_1 } = 26.62{\text{ m/s}} \hfill \\$

${\text{Put this value in eqn}}{\text{.(1)}} \hfill \\$

$\overrightarrow {v_2 } = 26.62 - 23.52 \hfill \\$

$\overrightarrow {v_2 } = 3.1{\text{ m/s}} \hfill \\$

${\text{So, the speed of the car is 3}}{\text{.1 m/s, when it hits the tree}}{\text{.}} \hfill \\$
• September 8th 2008, 07:48 PM
2clients
Quote:

Originally Posted by Shyam
${\text{Now, }}\overrightarrow a = {\text{acceleration}} = - 5.6{\text{ m/s}}^2 \hfill \\$

$\overrightarrow {v_1 } = {\text{ initial velocity of car}} \hfill \\$

$\overrightarrow {v_2 } = {\text{ final velocity of car}} \hfill \\$

$\Delta t = {\text{ time interval }} = 4.2{\text{ s}} \hfill \\$

$\Delta \overrightarrow d = {\text{ displacement of car }} = 62.4{\text{ m}}{\text{.}} \hfill \\$

${\text{Now, using the formula, Acceleration, }}\overrightarrow a = \frac{{\overrightarrow {v_2 } - \overrightarrow {v_1 } }}
{{\Delta t}} \hfill \\$

$\overrightarrow {v_2 } - \overrightarrow {v_1 } = - 5.6 \times 4.20 \hfill \\$

$\overrightarrow {v_2 } = \overrightarrow {v_1 } - 23.52{\text{ }}......................{\text{eqn}}{\text{. (1)}} \hfill \\$

${\text{Now, Displacement, }}\Delta \overrightarrow d = \overrightarrow {v_1 } \Delta t + \frac{1}
{2}\overrightarrow a \left( {\Delta t} \right)^2 \hfill \\$

$62.4 = \overrightarrow {v_1 } \left( {4.2} \right) + \frac{1}
{2}\left( { - 5.6} \right)\left( {4.2} \right)^2 \hfill \\$

$\overrightarrow {v_1 } = 26.62{\text{ m/s}} \hfill \\$

${\text{Put this value in eqn}}{\text{.(1)}} \hfill \\$

$\overrightarrow {v_2 } = 26.62 - 23.52 \hfill \\$

$\overrightarrow {v_2 } = 3.1{\text{ m/s}} \hfill \\$

${\text{So, the speed of the car is 3}}{\text{.1 m/s, when it hits the tree}}{\text{.}} \hfill \\$

Thank you very much!!! It makes perfect sense now.
• September 9th 2008, 06:37 PM
skeeter
derivation of $\Delta x = v_f t - \frac{1}{2}at^2$ ...

$\Delta x = v_{avg} t$

for constant a, $v_{avg} = \frac{v_f + v_0}{2}$

$\Delta x = \frac{v_f + v_0}{2}t$

$\Delta x = \frac{v_f}{2}t + \frac{v_0}{2}t$

$\Delta x = v_f t - \frac{v_f}{2}t + \frac{v_0}{2}t$

$\Delta x = v_f t - \frac{v_f - v_0}{2}t$

$\Delta x = v_f t - \frac{1}{2} \cdot \frac{v_f - v_0}{t}t^2$

$\Delta x = v_f t - \frac{1}{2}at^2$

$v_f = \frac{\Delta x}{t} + \frac{1}{2}at = \frac{62.4}{4.2} - (2.8)(4.2) = 3.1 \, m/s$