# A man walks then runs...

• June 20th 2009, 05:49 AM
Stroodle
A man walks then runs...
A man walks at a speed of 2 km/h for 45 minutes and then runs at 4 km/h for 30 minutes. Let $S$ km be the distance the man has run after $t$ minutes. The distance travelled can be described by:

$S(t)=\left\{\begin{array}{ll}at &\mbox{ if } 0\leq t\leq c\\bt+d, & \mbox{ if } c

Find the values of $a,\;b,\;c,\;d\;$ and $\;e$

I can't get my head around this question. All I have so far is $c$=45 and $e$=75

• June 20th 2009, 06:08 AM
yeongil
Quote:

Originally Posted by Stroodle
A man walks at a speed of 2 km/h for 45 minutes and then runs at 4 km/h for 30 minutes. Let $S$ km be the distance the man has run after $t$ minutes. The distance travelled can be described by:

$S(t)=\left\{\begin{array}{ll}at &\mbox{ if } 0\leq t\leq c\\bt+d, & \mbox{ if } c

Find the values of $a,\;b,\;c,\;d\;$ and $\;e$

I can't get my head around this question. All I have so far is $c$=45 and $e$=75

Remember: "distance equals rate times time", or d = rt.

a is going to be the first speed. But it's 2 km/h, and t is supposed to be in minutes, so convert: $2 \;\text{km/h} = \frac{1}{30}\;\text{km/min} = a$.

The second speed is going to be $4 \;\text{km/h} = \frac{1}{15}\;\text{km/min} = b$.

Note that the speed changes after 45 minutes. We need to know the distance traveled: $\left(\frac{1}{30}\;\text{km/min}\right)\times 45\;\text{min} = 1.5\;\text{km}$.

The equation find the distance after having run for 45 minutes @ 2 km/h would be
\begin{aligned}
\frac{1}{15}(t - 45) + 1.5 &= \frac{1}{15}t - 3 + 1.5 \\
&= \frac{1}{15}t - 1.5 \\
&= bt + d
\end{aligned}

so b = 1/15 and d = -1.5.

Therefore,
$S(t)=\left\{\begin{array}{ll}\frac{1}{30}t &\mbox{ if } 0\leq t\leq 45\\\frac{1}{15}t-1.5, & \mbox{ if } 45

01
• June 20th 2009, 06:17 AM
Stroodle
Haha. I just didn't notice that it was in minutes...
I spent like 20 mins wondering why a wasn't 2 :)
Thanks!