An office goer sees a bus trundling past at 10 km/h
and makes a run for it at 14 km/h. If he manages to
board the bus after 9 seconds, how far was the bus's
entrance when the office goer started running?
Answers
Answer:
We need to get everything in the same units first:
1hr=3600s
This gives: 10km/hr=10km/3600s=1km/360s
14km/hr=14km/3600s=7km/1800s
In 9 seconds the bus traveled 1km/360s X 9s = 9km/360=1km/40 or 0.025km.
In 9 seconds the person traveled 7km/1800s X 9s = 7km/200 or 0.035km.
Subtracting the 2 shows that the person had to run 0.01km farther to catch up, so the answer is 0.01km.
Step-by-step explanation:
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Maria S. asked • 07/20/18
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An office goers sees a bus trundling past at 10 km/hr and makes a run for it at 14 km/hr.If he manages to board the bus after 9 seconds ,how far was the bus's entrance when the office goer started running.
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David M. answered • 07/21/18
TUTOR 4.8 (122)
Dave "The Math Whiz"
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We need to get everything in the same units first:
1hr=3600s
This gives: 10km/hr=10km/3600s=1km/360s
14km/hr=14km/3600s=7km/1800s
In 9 seconds the bus traveled 1km/360s X 9s = 9km/360=1km/40 or 0.025km.
In 9 seconds the person traveled 7km/1800s X 9s = 7km/200 or 0.035km.
Subtracting the 2 shows that the person had to run 0.01km farther to catch up, so the answer is 0.01km.
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Andy C. answered • 07/20/18
TUTOR 4.9 (27)
Math/Physics Tutor
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T is the delay time
The bus travels for 9+T seconds at 10 km/hr
The office goer travels for 9 seconds at 14 km/hr
Next the speeds are changed to km/sec to match the times.
1 hour = 60 min and 1 min = 60 sec, so 1 hour = 3600 sec
So the bus speed is 1/360 km/sec
and the office goer speed is 14/3600 = 7/1800 km/sec
Their distances are the same:
(T+9)(1/360) = 9(7/1800)
(T+9)/360 = 7/200
200(T+9) = 360*7
T = 360*7 /200 - 9
= 3.6
So the bus went (1/360)(12.6) = 0.035 km = 35 meters
The office goer ran (7/1800)(9) = 7/200 km = 0.035 km = 35 meters
So in 3.6 seconds the bus was 3.6 * 1/360 km = 3.6/360 km = 36/3600 km = 1/100 km = 10 meters ahead
Answer:
We need to get everything in the same units first:
1hr=3600s
This gives: 10km/hr=10km/3600s=1km/360s
14km/hr=14km/3600s=7km/1800s
In 9 seconds the bus traveled 1km/360s X 9s = 9km/360=1km/40 or 0.025km.
In 9 seconds the person traveled 7km/1800s X 9s = 7km/200 or 0.035km.
Subtracting the 2 shows that the person had to run 0.01km farther to catch up, so the answer is 0.01km.
T is the delay time
The bus travels for 9+T seconds at 10 km/hr
The office goer travels for 9 seconds at 14 km/hr
Next the speeds are changed to km/sec to match the times.
1 hour = 60 min and 1 min = 60 sec, so 1 hour = 3600 sec
So the bus speed is 1/360 km/sec
and the office goer speed is 14/3600 = 7/1800 km/sec
Their distances are the same:
(T+9)(1/360) = 9(7/1800)
(T+9)/360 = 7/200
200(T+9) = 360*7
T = 360*7 /200 - 9
= 3.6
So the bus went (1/360)(12.6) = 0.035 km = 35 meters
The office goer ran (7/1800)(9) = 7/200 km = 0.035 km = 35 meters
So in 3.6 seconds the bus was 3.6 * 1/360 km = 3.6/360 km = 36/3600 km = 1/100 km = 10 meters ahead