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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+20585 
+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
 #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+20585 
+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

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