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
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.
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.
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.