A car travels from one town to another at a speed of 32mph. If it had gone 4 mph faster, it could have made the trip in 1/2 hr less time. How far apart are the towns?
So far I have rate of 32mph, 32+4
x = initial time
y = x-1/2
Please help.
A car travels from one town to another at a speed of 32mph. If it had gone 4 mph faster, it could have made the trip in 1/2 hr less time. How far apart are the towns?
So far I have rate of 32mph, 32+4
x = initial time
y = x-1/2
Please help.
speed= distance/time so, letting d be the distance between the two towns, in miles, and t be the time, in hours, for the trip, 32= d/t.
If "it had gone 4mph faster" its speed would have been 32+ 4= 36 and "it could have made the trip in 1/2 hr less time" so the time would be t- 1/2. Now "speed= distance/time" so 36= d/(t+1/2).
32= d/t is the same as d= 32t.
36= d/(t+1/2) is the same as d= 36(t+1/2)= 36t+ 18.
d= 32t= 36t+ 18.
Solve 32t= 36t+ 18 for t and then use d= 32t to find the distance between the towns.
Hmm that looks overly confusing, mine was like:
$\displaystyle 32 = \frac{d}{t}$
$\displaystyle d = 32t$
$\displaystyle 36 = \frac{d}{t-\frac{1}{2}} $
$\displaystyle d = 36(t-\frac{1}{2}) = 36t - 18 $
$\displaystyle \therefore 32t = 36t - 18 $
$\displaystyle -4t = -18 $
$\displaystyle t = \frac{-18}{-4} = 4\frac{1}{2}$
Then obviously sub into d=32t like you have done to get 144.
Does not simplify to 4t = 18, the -32t on each side would cancel to leave 36t = 18 so you've gone wrong somewhere.36t - 32t = -32t + 18