QR Question (Rate & Distance)

[suns6233]

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two distances problem I am not able to get the right answer. Anyone can try it

1) Mary leaves for a trip driving at 52 miles/hr. One-half hour later, Beth leaves on the same interstate highway traveling 62 miles/hr. How many miles will Beth have to drive before she overtakes Mary?

here is what I did: let t= Mary's time
(t-1.5)= Beth time
(t*52) = (t-1.5)*62
I figured the time and subtracted 1.5 from it; thus, t-1.5 which is for Beth
Then, used D= r*t
but did not get the right answer, which is 161.2 miles.


2)Mike leaves his house for Bentleyville, which is 200 miles away. After 3 hrs, he stops for lunch for 30 minutes; then he drives 10 miles slower for the rest of the trip. If the trip takes 5 hrs, what was his beginning speed?

This one took a little bit to figure out how to set up the problem, but then don't know how to get the answer.
Right answer is: 47.8 miles
anyone who can help is appreciated.

 

[Lnguyen7]

here is what I did: let t= Mary's time
(t-1.5)= Beth time
(t*52) = (t-1.5)*62

It should be
(t-0.5)= Beth time
(t*52) = (t-0.5)*62

2) Let v1 be his initial speed and v2 be his later speed, so v2 = v1 - 10 (1)
v2 = d2/t2 or d2 = v2*t2 (2)
v1 = d1/t1 (3)
The total trip is 200 miles or d1 + d2 = 200 or d1 = 200 - d2 (4)
Subtitute (4) into (3): v1 = d1/t1 = (200 - d2)/t1 (5)
Subtitute (2) into (5): v1 = d1/t1 = (200 - d2)/t1 = (200 - v2*t2)/t1 (6)
Subtitute (1) into (6): v1 = d1/t1 = (200 - d2)/t1 = (200 - v2*t2)/t1 = (200 - [v1 - 10]t2)/t1

t1 = 3 hours
t2 = 5 - 3 - 0.5 = 1.5 hours
So v1 = (200 - [v1 - 10]1.5)/3 = 47.8 miles/hr

 

[DAT Destroyer]

two distances problem I am not able to get the right answer. Anyone can try it

1) Mary leaves for a trip driving at 52 miles/hr. One-half hour later, Beth leaves on the same interstate highway traveling 62 miles/hr. How many miles will Beth have to drive before she overtakes Mary?

here is what I did: let t= Mary's time
(t-1.5)= Beth time
(t*52) = (t-1.5)*62
I figured the time and subtracted 1.5 from it; thus, t-1.5 which is for Beth
Then, used D= r*t
but did not get the right answer, which is 161.2 miles.


2)Mike leaves his house for Bentleyville, which is 200 miles away. After 3 hrs, he stops for lunch for 30 minutes; then he drives 10 miles slower for the rest of the trip. If the trip takes 5 hrs, what was his beginning speed?

This one took a little bit to figure out how to set up the problem, but then don't know how to get the answer.
Right answer is: 47.8 miles
anyone who can help is appreciated.

For the first problem:
After 30 min Mary has travelled 26 miles. ( 52*0.5=26). When Beth starts she's 26 miles behind Mary. When Beth catches up with Mary we have the following:
Db = Dm + 26. That is the distance traveled by Beth is the distance traveled by Mary plus 26 miles.
62*t = 52*t + 26. Solve for t to get : t=2.6

To find the total distance Beth travelled:
Db = 26*2.6 = 161.2 miles.

Second problem:
Total distance = distance traveled in 3 hours plus the distance traveled in 1.5 hours.

200 = 3v + 1.5( v - 10) solve for v to get: 47.8 mph
Note: t = 1.5 hours for the second half of the trip because the entire trip took 5 hours and Mike took a 30 min break.

Hope this helps