SOLUTION: a motorist travelling at constant speed leaves A at 11 00 intending to arrive B , 100km away at 1300 . half an hour later one of the tyres are punctured ands he is delayed for 18 m

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Question 1134823: a motorist travelling at constant speed leaves A at 11 00 intending to arrive B , 100km away at 1300 . half an hour later one of the tyres are punctured ands he is delayed for 18 mins . how fast must he then proceed in order to reach B on time ? at what time will he meet a cyclist who leaves town B at 11 45 for A ., travelling at a contant speed of 20 km/h ? make a distancwe time graph
Found 2 solutions by josmiceli, MathTherapy:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s%5B1%5D+ = motorist's initial constant speed in km/hr
In 1/2 hr, he travels +s%5B1%5D%2A%281%2F2%29+ km
--------------
He needs to make deadline of 2 hrs to get to B
After the 1/2 hr of travel, he loses 18 min, so now
he needs to get to B from location of puncture
in +2+-+1%2F2+-+18%2F60+ hrs
--------------------------------
The distance he has to travel is +100+-+s%5B1%5D%2A%281%2F2%29+
I can say:
+100+-+.5s%5B1%5D+=+s%5B2%5D%2A%28+1.5+-+.3+%29+
+100+-+.5s%5B1%5D+=+1.2s%5B2%5D+
+s%5B2%5D+=+%28+100+-+.5s%5B1%5D+%29+%2F+1.2+
--------------------------------------
Looking at the total time, I can say:
+2+=+100%2Fs%5B1%5D+%2B+1.2+
+100%2Fs%5B1%5D+=+.8+
+s%5B1%5D+=+100%2F.8+
+s%5B1%5D+=+125+ km/hr
and
+s%5B2%5D+=+%28+100+-+.5%2A125+%29+%2F+1.2+
+s%5B2%5D+=+%28+100+-+62.5+%29+%2F+1.2+
+s%5B2%5D+=+37.5%2F1.2+
+s%5B2%5D+=+31.25+ km/hr answer
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check answer:
+100+-+.5s%5B1%5D+=+1.2s%5B2%5D+
+100+-+.5%2A125+=+1.2%2A31.25+
+100+-+62.5+=+37.5+
+37.5+=+37.5+
----------------------
+s%5B1%5D%2A.5+=+125%2A.5+
+s%5B1%5D%2A.5+=+62.5+ km
and
+s%5B2%5D%2A1.2+=+31.25%2A1.2+
+s%5B2%5D%2A1.2+=+37.5+
+62.5+%2B+37.5+=+100+
OK
----------------------------
That's as far as I can go due to time




Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

a motorist travelling at constant speed leaves A at 11 00 intending to arrive B , 100km away at 1300 . half an hour later one of the tyres are punctured ands he is delayed for 18 mins . how fast must he then proceed in order to reach B on time ? at what time will he meet a cyclist who leaves town B at 11 45 for A ., travelling at a contant speed of 20 km/h ? make a distancwe time graph
The other person's answer of 31.25 km/h can't be correct and just doesn't make sense.

 

With a time of 2 (1300  -  1100) hours to cover a distance of 100 km, his planned average-speed was matrix%281%2C4%2C+100%2F2%2C+or%2C+50%2C+%22km%2Fh%22%29 for the trip

At matrix%281%2C6%2C+50%2C+%22km%2Fh%22%2C+and%2C+1%2F2%2C+hour%2C+later%29, distance covered was: matrix%281%2C6%2C+50%2C+%22%2A%22%2C+1%2F2%2C+%22=%22%2C+25%2C+km%29



matrix%281%2C8%2C+18%2C+minutes%2C+%22=%22%2C+18%2F60%2C+hour%2C+or%2C+.3%2C+hour%29

Since he has 1.2 (2 - .5  -  .3) hours to cover 75 (100  -  25) km, then his speed MUST be: highlight_green%28matrix%281%2C4%2C+75%2F1.2%2C+%22=%22%2C+62.5%2C+%22km%2Fh%22%29%29. 

**NOTE: He MUST travel FASTER than his planned average-speed in order to make it on time.