+0  
 
0
503
2
avatar

Tom and Diane start to race. Tom took 4 seconds to run 6 yards. Diane ran 5 yards in 3 seconds. If they continued to run at the same speeds, who would get to 30 yards first? Show how you figured it out please

Guest Mar 10, 2015

Best Answer 

 #2
avatar+18827 
+10

Tom and Diane start to race. Tom took 4 seconds to run 6 yards. Diane ran 5 yards in 3 seconds. If they continued to run at the same speeds, who would get to 30 yards first ?

you can get the equation for velocity as

$$v = \dfrac{d}{t}$$

Velocity (v) or speed equals the distance (d) traveled divided by the time (t) it takes to go that distance.

 

$$\\\small{\text{
$
\begin{array}{lc|cc}
\hline
\\
t_{Tom}= \dfrac{30\ yards}{v_{Tom}}
&
\qquad
&
\qquad
&
v_{Tom} = \dfrac{6\ yards}{4\ seconds} \\\\
t_{Tom}= \dfrac{30\ yards}{ \dfrac{6\ yards}{4\ seconds} }
=\dfrac{ 30\ yards \cdot 4\ seconds }{ 6\ yards }\\\\
t_{Tom}= 20 \ seconds
\end{array}
$}}\\\\
\begin{array}{lc|cc}
\hline
\\
t_{Diane}= \dfrac{30\ yards}{v_{Diane}}
&
\qquad
&
\qquad
&
v_{Diane} = \dfrac{5\ yards}{3\ seconds} \\\\
t_{Diane}= \dfrac{30\ yards}{ \dfrac{5\ yards}{3\ seconds} }
=\dfrac{ 30\ yards \cdot 3\ seconds }{ 5\ yards }\\\\
t_{Diane}= 18 \ seconds
\end{array}$$

Diane get  to 30 yards first in 18 seconds.

heureka  Mar 10, 2015
Sort: 

2+0 Answers

 #1
avatar
+5

speed = distance covered/time taken.

speed of Tom = 6 yards/4 seconds = 1.5 yards per second ,,

speed of Diane = 5 yards/3 seconds = 1.6667 yards per second ,,

since speed of Diane is much more than that of Tom ,,,,, 

DIANE WILL GET TO 30 YARDS FIRST.

Guest Mar 10, 2015
 #2
avatar+18827 
+10
Best Answer

Tom and Diane start to race. Tom took 4 seconds to run 6 yards. Diane ran 5 yards in 3 seconds. If they continued to run at the same speeds, who would get to 30 yards first ?

you can get the equation for velocity as

$$v = \dfrac{d}{t}$$

Velocity (v) or speed equals the distance (d) traveled divided by the time (t) it takes to go that distance.

 

$$\\\small{\text{
$
\begin{array}{lc|cc}
\hline
\\
t_{Tom}= \dfrac{30\ yards}{v_{Tom}}
&
\qquad
&
\qquad
&
v_{Tom} = \dfrac{6\ yards}{4\ seconds} \\\\
t_{Tom}= \dfrac{30\ yards}{ \dfrac{6\ yards}{4\ seconds} }
=\dfrac{ 30\ yards \cdot 4\ seconds }{ 6\ yards }\\\\
t_{Tom}= 20 \ seconds
\end{array}
$}}\\\\
\begin{array}{lc|cc}
\hline
\\
t_{Diane}= \dfrac{30\ yards}{v_{Diane}}
&
\qquad
&
\qquad
&
v_{Diane} = \dfrac{5\ yards}{3\ seconds} \\\\
t_{Diane}= \dfrac{30\ yards}{ \dfrac{5\ yards}{3\ seconds} }
=\dfrac{ 30\ yards \cdot 3\ seconds }{ 5\ yards }\\\\
t_{Diane}= 18 \ seconds
\end{array}$$

Diane get  to 30 yards first in 18 seconds.

heureka  Mar 10, 2015

18 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details