# Really easy distance/time problem I'm having a brain freeze on

1. Aug 11, 2009

### Paulo Serrano

1. The problem statement, all variables and given/known data
Driving to a nearby city, a motorist estimates his travel time, considering that he can maintain a constant velocity of 90km/h. Surprised by an unexpected rain shower, he decides to reduce his velocity to 60km/h until it stops raining 15 minutes later, at which time he returns to his original velocity. This temporary reduction in speed increases his travel time, in relation to his original estimate by how much:
a) 5 minutes
b) 7.5 minutes
c) 10 minutes
d) 15 minutes
e) 30 minutes

2. Relevant equations

D=vt (I think that's all really)

3. The attempt at a solution

I know that, because of the rain, the driver drove 15 km at the reduced speed of 60km/h (which was also for a duration of 15 minutes) but when I actually try to do the problem I can't seem to get the right answer. The funny thing is I've done this question before and got the correct answer easily.

2. Aug 11, 2009

### Staff: Mentor

One intuitive way to help you get your head around the problem is to draw a d(t) graph, with distance travelled on the vertical axis, and time on the horizontal axis. Draw the start of the trip as a line going up to the right with the slope corresponding to 90kph. At some point, reduce the slope to 60kph for 15 minutes, and then resume the original 90kph slope. Now draw a horizontal line at some distance that is after the resumption of 90kph. See how that horizontal line intersects the two lines (original 90kph-only line and the modified line) at two places? The difference in time between those two intersection points stays constant as time and distance proceed after returning to the original speed...

3. Aug 11, 2009

### ideasrule

So he drove 30 km/h slower for 15 minutes. A hypothetical car driving 90 km/h would see him receding at 30 km/h for 15 minutes, and get left behind by a certain distance (what's the distance?) He then has to cover that distance at 90 km/h.

Another intuitive way of doing it is to calculate the time he would take at 90 km/h, calculate the time he actually took, and subtract the two. Since you don't know the distance, call it "x" and hope that it cancels out at the end.

4. Aug 11, 2009

### kuruman

Yet another way:

At 90 km/hr, he covers 1.5 miles/min (90/60)
At 60 km/hr, he covers 1.0 miles/min

So over the 15 minutes that it's raining, he falls behind 0.5*15 = 7.5 miles.
How long will it take him to cover that "lag distance" if he actually traveled at 1.5 miles/min? That's the lag time.

5. Aug 11, 2009

### Staff: Mentor

Okay, that's enough help everybody. Now it's time to let the OP finish off the problem.

And we'll see if he can spot the typo in the quoted post above...

6. Aug 11, 2009

### kuruman

Oops. I just got back from a driving (US) trip.

7. Aug 12, 2009

### Paulo Serrano

I knew it was easy. :p

Using a slight variation of berkeman's method, I found that the answer is A, 5 minutes.

90 km/h ---------> 1.5 km/min
60 km/h ---------> 1 km/min

The distance traveled at the reduced speed was 15 km. How long does it take for each speed?
1.5 km/min -----> 10 minutes
1.0 km/min -----> 15 minutes

The difference between the two speeds is 5 minutes.

Thanks guys.