Stop Race Car in 1000 m with Parachute: Solve for Retarding Force

[karush]

The parachute on a race car of weight 8820 N opens at the end of quater mile run when the car is traveling at 35m/s^2
a.What total retarding force must be supplied by the parachute to stop the car to a distance of 1000 m

$\displaystyle d = v_0\cdot t + \frac{1}{2}\cdot a \cdot t^2 $
$\displaystyle v = v_0 + a\cdot t $ok I think you use these 2 equations
but not sure how to do the steps

 

[MarkFL]

We are asked to find the magnitude of a force, and Newton's 2nd LAw states essentially:

$$F=ma$$

We are given the mass, the displacement, the initial and final velocities and so I would look at the following kinematic equation:

$$a=\frac{v_f^2-v_i^2}{2x}$$

Can you put these two equations together to find the required retarding force?

 

[MarkFL]

what is x ?

1000?

Yes, $x$ is the displacement, which is given as 1000 m. Since this is physics, I would recommend always using units with your numbers, and ensure your units work out correctly. :)

 

[karush]

Yes, $x$ is the displacement, which is given as 1000 m. Since this is physics, I would recommend always using units with your numbers, and ensure your units work out correctly. :)

this?

$\displaystyle a=\frac{v_f^2-v_i^2}{2x}
\implies\frac{0^2-(35 m/s^2)^2}{2(1000 m)}$

then?

$F=8820a$

 

[MarkFL]

$\displaystyle a=\frac{v_f^2-v_i^2}{2x}
\implies\frac{0^2-(35 m/s^2)^2}{2(1000 m)}$

this?

$\displaystyle a=\frac{v_f^2-v_i^2}{2x}
\implies\frac{0^2-(35 m/s^2)^2}{2(1000 m)}$

then?

$F=8820a$

You've got the right idea, but your units are off...I would write:

$$|F|=m|a|=m\left|\frac{v_f^2-v_i^2}{2x}\right|$$

Okay, now given that $v_f^2<v_i^2$, we can write:

$$|F|=\frac{m\left(v_i^2-v_f^2\right)}{2x}$$

Now, we are given the weight of the car, but we need the mass:

$$m=\frac{w}{g}$$

Hence:

$$|F|=\frac{w\left(v_i^2-v_f^2\right)}{2gx}$$

Now, plug in the given data along with the units...what do you get?

 

[karush]

Now, we are given the weight of the car, but we need the mass:

$$m=\frac{w}{g}$$

Hence:

$$|F|=\frac{8820\left(v_i^2-v_f^2\right)}{2gx}$$

Now, plug in the given data along with the units...what do you get?

$\displaystyle |F|=\Biggr|\frac{w\left(v_i^2-v_f^2\right)}{2gx}\Biggr|$
$\displaystyle |F|=\Biggr|\frac{8820((35 m/s^2)^2-0^2)}{2(9.8 m/s^2)(1000m)}\Biggr|$kinda?
kinda?

 

[MarkFL]

$\displaystyle |F|=\Biggr|\frac{w\left(v_i^2-v_f^2\right)}{2gx}\Biggr|$
$\displaystyle |F|=\Biggr|\frac{8820((35 m/s^2)^2-0^2)}{2(9.8 m/s^2)(1000m)}\Biggr|$kinda?
kinda?

Your units are off, or missing...what you want is:

$$|F|=\frac{\left(8820\text{ N}\right)\left(\left(35\dfrac{\text{m}}{\text{s}}\right)^2-\left(0\dfrac{\text{m}}{\text{s}}\right)^2\right)}{2\left(9.8\dfrac{\text{m}}{\text{s}^2}\right)\left(1000\text{ m}\right)}$$

Do you see how all the units divide out leaving nothing by Newtons, which is dimensionally correct, since the LHS is the magnitude of a force? So what we're left with is:

$$|F|=\frac{8820\cdot35^2}{2\cdot9.8\cdot1000}\,\text{N}=\,?$$

 

[karush]

$\displaystyle
|F|=\Biggr|\frac{\left(8820\text{ N}\right)\left(\left(35\dfrac{\text{m}}{\text{s}}\right)^2-\left(0\dfrac{\text{m}}{\text{s}}\right)^2\right)}
{2\left(9.8\dfrac{\text{m}}{\text{s}^2}\right)\left(1000\text{ m}\right)}\Biggr|
=\color{red}{551.25 \, \textsf{N}}$

the W|A touch!