distance

Jomo said:
It is good that you did not understand. I was assuming he went to the event at 45mph and returned home at 75 mph.
So forget that.
Let t represents the time it took me to get to the event at the slower speed
Going to the event at 45mph: d=45t
Going to the event at 75 mph: d = 75(t-2)

45t = 75t - 150
-45t + 150 = -45t + 150
150 = 30t
t=5 hours.
Click to expand...
Why t-2?. Where does the 2 come from?.
 
Look at the definition of t!!!!!!!!!!!!!!!!!!!!!! And read lev888's post!
 
eddy2017 said:
Hi, dear teachers and friends:
If Robert drives to an event at an average speed of 45 miles per hour he will arrive an hour late: however if he drives at an average speed of 75 mph he will arrive an hour early. What is the distance that is driving to get to the event?.
As always appreciating no end your helpful hints and help.
eddy
Given
if Robert drives to an event at an average speed of 45 mph he will arrive 1 hour late.
if he drives at 75 mph h will arrive 1 hour early.
d?
Click to expand...
Find of the problem: What is the distance that is driving to get to the event?.

let the distance to drive = d

Let us assume:

the time (in hr.) taken to arrive at the event at 45 mph = t

Then the time taken to arrive at the event at 75 mph = (t -2)

Then we use the equation

distance = (time) * (speed) ................................................. (in this case distance remains the same at 45 mph or 75 mph

Then we get,

t * 45 = (t-2) * 75

45 * t = 75 * t - 150

150 = 75*t - 45*t

t = 5

d = t * 45 = 225 mile

check

d = (5-2) * 75 = 3 * 75 = 225 ........................................................... checks
 
Jomo, at post #54 you agreed that the time was 4 hours.
Then here at # 60 you write t=5
How come?.
Which is the correct one for you?.
I re-did it again and it gives me 4 hours. I did it again making a chart and it gives me 4 hours.
 
eddy2017 said:
Jomo, at post 354 you agreed that the time ws 4 hours.
Then here at # 60 you write t=5
Click to expand...
Yes, that is 100% true. Did you read the definitions of those two t's? They are defined differently. That is why they have different values. I showed you two different ways to solve the problem using different definitions for 4 so it is only reasonable for the values of t to be different.
 
eddy2017 said:
Jomo, at post #54 you agreed that the time was 4 hours.
Then here at # 60 you write t=5
How come?.
Which is the correct one for you?.
I re-did it again and it gives me 4 hors.
Click to expand...
The problem references 3 different times:
1. Too early
2. "Correct time"
3. Too late
That's why it's important to define the variables exactly. (edit: and to read someone else's definitions carefully)
 
I will draw a chart and fill in the rates, and i will let 75(t-1) be when early.
45(t+1) when late.
rt(rate time) be when on time.
I will continue. I will post the whole thing.
 
Here is what I hope is a simple problem which I will do twice, get different answers for x but the end result has the same numbers.

Three consecutive integers add up to 24. Find the three numbers
Let x = smallest number
Let x+1 = middle sized number
Let x+2 = largest sized number
Adding all three we get (x) + (x+1) + (x+2) = 24.
3x + 3 = 24
3x = 21
x=7
So x=7 is the smallest number.
x+1 = 7+1 = 8 = middle sized number
x+2 = 7+2 = 9 = largest sized number.

Three consecutive integers add up to 24. Find the three numbers
let x-1 = smallest number
let x = middle sized number
let x+1 = largest sized number
Adding all three we get (x-1) + (x) + (x+1) = 24
3x=24
x=8
x-1 = 8-1 = 7 = smallest number
x=8 = middle sized number
x+1 = 8 + 1= 9 = largest sized number


In both cases we got the three numbers are 7, 8 and 9. That is we got the same answers! However we got different values for x! One x ended up equaling 7 and another x ended up x equaling 8. Why? They were defined differently.

Just not to confuse this thread please do not ask questions regarding this post on this thread here. If you need to ask any questions regarding this problem then please start a new post.
 
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Chart
distance rate time
when early 75 t-1
when late 45 t+1
when on time r t

Now, i will ise the formula to find distance
d=rt and fill in the distances

distance rate time
when early 75(t-1) 75 t-1
when late 45(t+1) 45 t+1
when on time rt r t

The three distances are equal. They are. it is the time to get there and the speeds s so far the only thing that is different.

I will set the first two distances equal.
75(t-1) = 45(t+1)
75t-75= 45t+45
I will group like terms keeping both sides of the equation.
75t-45t=45+75
30t=120
t=4
the time to be on time is 4 hours.
What do you think of this?.
And if this is ok, what do i need to do next, because this is the time taken, but how to find the distance?
I know, d=rt.
 
Jomo said:
Here is what I hope is a simple problem which I will do twice, get different answers for x but the end result has the same numbers.

Three consecutive integers add up to 24. Find the three numbers
Let x = smallest number
Let x+1 = middle sized number
Let x+2 = largest sized number
Adding all three we get (x) + (x+1) + (x+2) = 24.
3x + 3 = 24
3x = 21
x=7
So x=7 is the smallest number.
x+1 = 7+1 = 8 = middle sized number
x+2 = 7+2 = 9 = largest sized number.

Three consecutive integers add up to 24. Find the three numbers
let x-1 = smallest number
let x = middle sized number
let x+1 = largest sized number
Adding all three we get (x-1) + (x) + (x+1) = 24
3x=24
x=8
x-1 = 8-1 = 7 = smallest number
x=8 = middle sized number
x+1 = 8 + 1= 9 = largest sized number


In both cases we got the three numbers are 7, 8 and 9. That is we got the same answers! However we got different values for x! One x ended up equaling 7 and another x ended up x equaling 8. Why? They were defined differently.

Just not to confuse this thread please do not ask questions on this post here. If you need to ask any questions regarding this problem then please start a new post.
Click to expand...
Oh, now i see, different values but same result. I see now. Thanks. How to proceed to find the distance?.
 
eddy2017 said:
Chart
distance rate time
when early 75 t-1
when late 45 t+1
when on time r t

Now, i will ise the formula to find distance
d=rt and fill in the distances

distance rate time
when early 75(t-1) 75 t-1
when late 45(t+1) 45 t+1
when on time rt r t

The three distances are equal. They are. it is the time to get there and the speeds s so far the only thing that is different.

I will set the first two distances equal.
75(t-1) = 45(t+1)
75t-75= 45t+45
I will group like terms keeping both sides of the equation.
75t-45t=45+75
30t=120
t=4
the time to be on time is 4 hours.
What do you think of this?.
And if this is ok, what do i need to do next, because this is the time taken, but how to find the distance?
I know, d=rt.
Click to expand...
You never answered the question! What is the distance! You claim that the distances are equal but do just state what the distance is. Please finish up.
 
75(t-1)
=75(4-1)
=75*3
d=225

45(4+1)
45*5
d= 225

We also know now that it should take him 4 hours to get to work if he drove at a speed that would get him there right on time.
 
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