Jason and Britton are driving to St. George. Jason got a 20 mile head start and drove an
1 answer:
Answer:
2 hours, 150 miles
Step-by-step explanation:
The relation between time, speed, and distance can be used to solve this problem. It can work well to consider just the distance between the drivers, and the speed at which that is changing.
<h3>Separation distance</h3>Jason got a head start of 20 miles, so that is the initial separation between the two drivers.
<h3>Closure speed</h3>Jason is driving 10 mph faster than Britton, so is closing the initial separation gap at that rate.
<h3>Closure time</h3>The relevant relation is ...
time = distance/speed
Then the time it takes to reduce the separation distance to zero is ...
closure time = separation distance / closure speed = 20 mi / (10 mi/h)
closure time = 2 h
Britton will catch up to Jason after 2 hours. In that time, Britton will have driven (2 h)(75 mi/h) = 150 miles.
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<em>Additional comment</em>
The attached graph shows the distance driven as a function of time from when Britton started. The distances will be equal after 2 hours, meaning the drivers are in the same place, 150 miles from their starting spot.
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Answer:
5
Step-by-step explanation:
for the first gear
revolutions/teeth
1 / 24
2 / 48
3 / 72
4 / 96
5 / 120
6 / 144
for the second gear
revolutions/teeth
1 / 40
2 / 80
3 / 120
4 / 160
<em>the two marks will meet after 120 teeth, 5 revolutions of the first gear and 3 revolutions of the second.</em>
the way to get that amount of teeth is
the Least Common Multiple equals the product of all factors, but those factors who are repeted for both numbers should be only once.
120 teeth are 5 revolutions for gear1 and 3 por gear2