I'm sure there was an easier way to do this

atwana_b

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Sep 2, 2006
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A train travels 20 feet in 1/5 second. At this same speed, how many feet will it travel in three seconds.

This is what I did

1/5 + 1/5 + 1/5 + 1/5 + 1/5 (because I know that I would need 5 to make a whole second) = 20 x 5 = 100

I actually did the above two more times and reached my answer of 300.

Help, I am sure that there was an easier way to do this.
 
d = rt

(distance = rate * time)

d = 20 feet
r = ?
t = (1/5) second

manipulate d=rt so that you get an equation for R: r = d/t

r = 20/(1/5)
r = 100 feet/second

now, use D(t)=RT

D(3) = 100(3)
D(3) = 300

There is your answer

D(3) means "the value of D when T=3"
 
atwana_b said:
A train travels 20 feet in 1/5 second. At this same speed, how many feet will it travel in three seconds.
1/5 sec = 20
1 sec = 20 * 5 = 100
3 sec = 300
 
Hello, atwana_b!

Your reasoning is correct.
That kind of logic is a Good Thing!


Here's another intuitive approach: . . . . . \(\displaystyle \frac{1}{5}\text{ second}\:\rightarrow\: 20\text{ feet}\)

. . . . . . Multiply both sides by 15: \(\displaystyle \;15\,\times\,\left(\frac{1}{5}\text{ second}\right) \;\rightarrow\;15\,\times\,(20\text{ feet})\)

. . . . . . . . . . . . . . . . and we get: . . . . \(\displaystyle 3\text{ seconds}\;\rightarrow\;300\text{ feet}\)
 
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