# Math Help - A man walks then runs...

1. ## 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

Can you please show working.
Thanks for your help

2. 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

Can you please show working.
Thanks for your help
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

3. Haha. I just didn't notice that it was in minutes...
I spent like 20 mins wondering why a wasn't 2
Thanks!
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.936636745929718, "perplexity": 845.2919104589614}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049276964.14/warc/CC-MAIN-20160524002116-00019-ip-10-185-217-139.ec2.internal.warc.gz"}