Word problem about distance, rate and time

  • Thread starter paulmdrdo
  • Start date
  • #1
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Homework Statement



Q35hall_XXIV.jpg

Homework Equations


D = rt

The Attempt at a Solution



I let d to be the total distance

d/3a = t1 ----- time taken for 1st part of the trip

d/3b = t2 ----- time taken for 2nd part of the trip

d/3c = t3 ------ time taken for the whole trip with uniform speed

Since t1+t2 =t3

d/3a + d/3b = d/3c

Multiplying both sides by 3abc

bcd + acd = abd

Dividing both sides by d

bc + ac = ab

Factoring out c and diving both sides by it.

1/c = (a + b)/ab = 1/a + 1/b

1/c = 1/a + 1/b not the same as 2/c =1/a + 1/b

Do you think it is just a typo?
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Answers and Replies

  • #2
Buzz Bloom
Gold Member
2,477
452
d/3c = t3 ------ time taken for the whole trip with uniform speed
Hi Paul:

The problems statement says:
"He could have ridden from A to B and back again in the same time."​
Therefore if t3 = time to go from A to B, then t3 = ( t1 + t2) / 2.

Regards,
Buzz
 
  • #3
ehild
Homework Helper
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1,915
2/c = 1/a +1/b is a typo. It should be 2/c = 1/a +2/b.
 
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
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1,124
Do you think it is just a typo?
No, it's not a typo.

See what Buzz said.
 

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