# Distance word problem

• February 20th 2009, 11:26 AM
aliveiam
Distance word problem
I am having trouble with this one. I am more interested in the steps than the answer.

A trip from Here to There is 2108 miles. If one car drove 6mph faster than a truck and got There 3 hours before the truck, find the speeds of the car and truck.

I think the car's speed is (truck's speed) + 6, its time is (Truck's time) - 3, and its distance is 2108, but I am unable to get an answer. I have manipulated the d=rt formula but to no avail. Any help would be appreciated.
• February 20th 2009, 11:50 AM
masters
Quote:

Originally Posted by aliveiam
I am having trouble with this one. I am more interested in the steps than the answer.

A trip from Here to There is 2108 miles. If one car drove 6mph faster than a truck and got There 3 hours before the truck, find the speeds of the car and truck.

I think the car's speed is (truck's speed) + 6, its time is (Truck's time) - 3, and its distance is 2108, but I am unable to get an answer. I have manipulated the d=rt formula but to no avail. Any help would be appreciated.

Hi aliveiam,

Let's see....There's probably an easier way to approach this, but this is what I saw first.

Let r = the truck's rate

Let t = the truck's time

Let r + 6 = the car's rate
Let t - 3 = the car's time (the car gets there 3 hours ahead of the truck)

D = 2108 for both of them

Using D = rt,

[1] ${\color{red}rt=2108}$

[2] ${\color{blue}(r+6)(t-3)=2108}$

Solve the system above.
• February 20th 2009, 11:52 AM
Rist
$\begin{cases}
Vt=2108 \\
(V+6)(t-3)=2108
\end{cases}$

Find V (truck's speed) and t (truck's time) out of it. Answers will be: truck's speed is 62, car's speed is 68.
• February 20th 2009, 12:16 PM
aliveiam
I am still having trouble
I appreciate your time with this problem, but I am still unable to come up with the correct solutions. When I begin with [r + 6][t - 3] = 2108, I want to solve for r so I change t in terms of r and I get [r + 6][(distance/rate) - 3] = 2108 which becomes [r + 6][(2108/(r + 6)) - 3] = 2108. When I solve for r, however, I get -6. Do you know where I went wrong? Thank you for your time.
• February 20th 2009, 12:31 PM
Rist
You're trying to solve it, using only one equation. Take t (2108/r) from the first one and put it in the second instead of finding t in the second one and putting it back there.
• February 20th 2009, 12:56 PM
masters
Quote:

Originally Posted by aliveiam
I appreciate your time with this problem, but I am still unable to come up with the correct solutions. When I begin with [r + 6][t - 3] = 2108, I want to solve for r so I change t in terms of r and I get [r + 6][(distance/rate) - 3] = 2108 which becomes [r + 6][(2108/(r + 6)) - 3] = 2108. When I solve for r, however, I get -6. Do you know where I went wrong? Thank you for your time.

Let me finish it up for you.

Quote:

Originally Posted by masters
Hi aliveiam,

Let's see....There's probably an easier way to approach this, but this is what I saw first.

Let r = the truck's rate

Let t = the truck's time

Let r + 6 = the car's rate
Let t - 3 = the car's time (the car gets there 3 hours ahead of the truck)

D = 2108 for both of them

Using D = rt,

[1] ${\color{red}rt=2108}$

[2] ${\color{blue}(r+6)(t-3)=2108}$

Solve the system above.

[2] ${\color{blue}(r+6)(t-3)=2108}$

$(r+6)(r-3)=2108$

$rt+6t-3r-18=2108$

Remember: [2] ${\color{red}rt=2108}$

$2108+6t-3r-18=2108$

[3] $6t-3r=18$

Now, use [1] ${\color{red}rt=2108}$ to solve for t in terms of r.

$t=\frac{2108}{r}$

Replace t in [3] $6t-3r=18$

$6\left(\frac{2108}{r}\right)-3r=18$

$\frac{12648}{r}-3r=18$

Multiply every term by r and rearrange to:

$3r^2+18r-12648=0$

Divide every term by 3.

$r^2+6r-4216=0$

Factor or use the quadratic formula:

$(r-62)(r+68)=0$

$r=62$

Ignore the negative root as it is extraneous (can't have a negative rate)

So the truck goes 62 mph and the car goes 62 + 6 = 68 mph.
• February 20th 2009, 01:08 PM
aliveiam
Thank you much
I see where I went wrong. Thank you for your support. All is well.