# Thread: Distance Time word problem

1. ## Distance Time word problem

Hi Guys,

I am having trouble figuring out this distance time word problem. I created a table and show my reasoning below but I am getting some weird answers. Can you guys please point out where I am going wrong? Thanks.

A car leaves town X at 0800 and arrives at town Y at 0915. During the first hour of the journey, the car travels at a constant speed of 80 km/h. If the average speed of the car between 0815 and 0915 is 2 km/h less than the average speed for the whole journey, find
(a) the distance of the journey from town X to town Y,
(b) the average speed between 0815 and 0915.

Here's my reasoning. I figured that if I assume x is the speed between 0900 to 0915 and then using the condition part of 2nd sentence, I would be able to solve for x. Here's the table I built,

$
\begin{tabular}{| l | c | c | c |}
Duration & Speed(km/h) & Time(h) & Distance(km) \\
0800-0900 & 80 & 1 & 80 \\
0800-0815 & 80 & 1/4 & 20 \\
0815-0900 & 80 & 3/4 & 60 \\
0900-0915 & x & 1/4 & x/4 \\
0815-0915 & 60 + x/4 & 1 & 60 + x/4 \\
\end{tabular}
$

Then, Avg Speed is

$
\begin{math}
\frac{80 + (60 + \frac{x}{4})}{\frac{5}{4}} = \frac{560 + x}{5}
\end{math}
$

And given,

$
\frac{560 + x}{5} - \frac{60+x}{4} = 2
$

Solving that equation gives me x = 1900 km/h!!!

P.S. First time using this Latex stuff , Thanks to great tutorial here.

2. Originally Posted by mathguy80
Hi Guys,

I am having trouble figuring out this distance time word problem. I created a table and show my reasoning below but I am getting some weird answers. Can you guys please point out where I am going wrong? Thanks.

A car leaves town X at 0800 and arrives at town Y at 0915. During the first hour of the journey, the car travels at a constant speed of 80 km/h. If the average speed of the car between 0815 and 0915 is 2 km/h less than the average speed for the whole journey, find
(a) the distance of the journey from town X to town Y,
(b) the average speed between 0815 and 0915.

Here's my reasoning. I figured that if I assume x is the speed between 0900 to 0915 and then using the condition part of 2nd sentence, I would be able to solve for x. Here's the table I built,

...
1. Draw a sketch (see attachment).

2. Let x denote the distance which is traveled between 0900 and 0915. Then the average speed of the whole trip is:

$v = \dfrac{80+x}{\frac54}$

3. The average speed between 0815 and 0915 is $2\ \frac{km}h$ less than v:

$\dfrac{60+x}1=\dfrac{80+x}{\frac54}-2$

4. Solve for x.

3. Hi @earboth

Thanks for your help. Sorry had some internet problems and wasn't able to get back to this sooner.

I tried it your way using x as distance. And it solves correctly! I am wondering if there is any particular reason you thought of using distance instead of speed as the variable? Using distance solves much cleaner.

Thanks.