# distance and time word problem

• Feb 10th 2009, 04:59 PM
vance
distance and time word problem
A man walks for 45 minutes at a rate of 3 mph, then jogs for 75 minutes at a rate of 5 mph, then sits and rests for 30 minutes, and finally walks for 1 hour and half.Find the rule of the function that expresses this distance traveled as a function of time [ Caution: Don't mix up the units of time ; use either minutes or hours , not both]

could someone help me with this problem? I sketched the graph ,and the ranges I got are from 0 to .75 , .75 to 2, 2 to 2.5 and 2.5 to 4 . I know that the first function is 3t and the third one is 6.25.However, I don't know how to write the other two ones.

• Feb 10th 2009, 06:28 PM
mollymcf2009
Quote:

Originally Posted by vance
A man walks for 45 minutes at a rate of 3 mph, then jogs for 75 minutes at a rate of 5 mph, then sits and rests for 30 minutes, and finally walks for 1 hour and half.Find the rule of the function that expresses this distance traveled as a function of time [ Caution: Don't mix up the units of time ; use either minutes or hours , not both]

could someone help me with this problem? I sketched the graph ,and the ranges I got are from 0 to .75 , .75 to 2, 2 to 2.5 and 2.5 to 4 . I know that the first rule is 3t and the third one is 6.25.However, I don't know how to write the other two ones.

Hello!

The function rule for this problem is
distance = rate x time
d = rt
To write the rule as the distance traveled as a function of the time it took to travel that distance you will need to evaluate each set of data using this function. Since you have 3 different rates of speed, and 4 time periods, you will have to calculate each one.

I changed the times to hours.

RATE TIME
0 $\displaystyle \rightarrow$ .5
3 $\displaystyle \rightarrow$ .75
5 $\displaystyle \rightarrow$ 1.25
3 $\displaystyle \rightarrow$ 1.5

So, distance as a function of time:
d= rt

1) $\displaystyle d(t) = 0t$ so $\displaystyle d(.5) = (0)(.5) = 0 miles$
2) $\displaystyle d(t) = 3t$ so $\displaystyle d(.75) = (3)(.75) = 2.25 miles$
3) $\displaystyle d(t) = 5(t)$ so $\displaystyle d(1.25) = (5)(1.25) = 6.25 miles$
4) $\displaystyle d(t) = 3t$ so $\displaystyle d(1.5) = (3)(1.5) = 4.5 miles$

Hope this is what you needed! Good luck! (Wink)
• Feb 10th 2009, 07:05 PM
Soroban
Hello, vance!

This is a tricky one!
Luckily, I've done one of these before . . .

Quote:

A man walks for 45 minutes at a rate of 3 mph, then jogs for 75 minutes at 5 mph,
then sits and rests for 30 minutes, and finally walks for 1 hour and half.
Find the rule of the function that expresses this distance traveled as a function of time.

This is a piece-wise function.

The distance function depends on what time period is involved.

From $\displaystyle t=0\text{ to }t = 0.75$, the distance function is: .$\displaystyle 3t$
. . By the end of the 45 minutes, he has gone 2.25 miles.

From $\displaystyle t = 0.75\text{ to } t = 2$, the distance function is: .$\displaystyle 2.25 + 5t$
. . By the end of the 75 minutes, he has gone another 6.25 miles, a total of 8.5 miles.

From $\displaystyle t = 2\text{ to }t = 2.5$, the distance function is: .$\displaystyle 8.5 + 0t \:=\:8.5$
. . By the end of the 30 minutes, he hasn't moved; his total is still 8.5 miles.

From $\displaystyle t = 2.5\text{ to }4$, the distance function is: .$\displaystyle 8.5+3t$
. . By the end of the 90 minutes, he has gone another 4.5 miles, a total of 13 miles.

The function would be written like this:

. . $\displaystyle d(t) \;=\;\begin{Bmatrix}3t & & 0 \leq t \leq 0.75 \\ 2.25 + 5t & & 0.75 \leq t < 2 \\ 8.5 & & 2 \leq t \leq 2.5 \\ 8.5+3t & & 2.5 < t < 4 \end{Bmatrix}$

The graph would look like this:
Code:

      |       |                              *       |                            *  :       |                        *    :       |                      *        :       |              * * *          :       |            * :  :          :       |          *  :  :          :       |        *    :  :          :       |      *      :  :          :       |    *        :  :          :       |  *  :        :  :          :   - - * - - + - - - - + - + - - - - - + - -       |    ¾        2  2½          4       |
• Feb 10th 2009, 08:03 PM
vance
Quote:

Originally Posted by Soroban
Hello, vance!

This is a tricky one!
Luckily, I've done one of these before . . .

This is a piece-wise function.

The distance function depends on what time period is involved.

From $\displaystyle t=0\text{ to }t = 0.75$, the distance function is: .$\displaystyle 3t$
. . By the end of the 45 minutes, he has gone 2.25 miles.

From $\displaystyle t = 0.75\text{ to } t = 2$, the distance function is: .$\displaystyle 2.25 + 5t$
. . By the end of the 75 minutes, he has gone another 6.25 miles, a total of 8.5 miles.

From $\displaystyle t = 2\text{ to }t = 2.5$, the distance function is: .$\displaystyle 8.5 + 0t \:=\:8.5$
. . By the end of the 30 minutes, he hasn't moved; his total is still 8.5 miles.

From $\displaystyle t = 2.5\text{ to }4$, the distance function is: .$\displaystyle 8.5+3t$
. . By the end of the 90 minutes, he has gone another 4.5 miles, a total of 13 miles.

The function would be written like this:

. . $\displaystyle d(t) \;=\;\begin{Bmatrix}3t & & 0 \leq t \leq 0.75 \\ 2.25 + 5t & & 0.75 \leq t < 2 \\ 8.5 & & 2 \leq t \leq 2.5 \\ 8.5+3t & & 2.5 < t < 4 \end{Bmatrix}$

The graph would look like this:
Code:

      |       |                              *       |                            *  :       |                        *    :       |                      *        :       |              * * *          :       |            * :  :          :       |          *  :  :          :       |        *    :  :          :       |      *      :  :          :       |    *        :  :          :       |  *  :        :  :          :   - - * - - + - - - - + - + - - - - - + - -       |    ¾        2  2½          4       |

Thanks soroban. This is what I was looking for!!!(Clapping)