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aleksandrvk [35]
3 years ago
9

A truck left a station at 6:00am traveling at a rate of 45km/h. A van left the same place 2 hours later heading in the same dire

ction at a rate of 54km/h. At what time will the van catch up with the truck?
Mathematics
2 answers:
OverLord2011 [107]3 years ago
6 0

The Van would catch up with the truck at 6:00 pm

Speed is the ratio of distance travelled to time taken. It is given by:

Speed = distance / time

Let us say the van catch up with the truck after t hours at a distance d.

For the truck:

45 = d/t

d = 45t

For the Van:

54 = d / (t - 2)

d = 54t - 108

45t = 54t - 108

9t = 108

t = 12 hours

The Van would catch up with the truck at 6:00 pm

Find out more on speed at: brainly.com/question/3004254

netineya [11]3 years ago
5 0

Answer:

The Van would catch up with the truck at 6:00 pm

Speed is the ratio of distance travelled to time taken. It is given by:

Speed = distance / time

Let us say the van catch up with the truck after t hours at a distance d.

For the truck:

45 = d/t

d = 45t

For the Van:

54 = d / (t - 2)

d = 54t - 108

45t = 54t - 108

9t = 108

t = 12 hours

The Van would catch up with the truck at 6:00 pm

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<h2>Answer:</h2>

To solve this problem we will use Heron's formula:

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Also:

(I) \ PS=SQ \\ \\ (II) \ SQ-PQ = 1 \\ \\ (III) \ PS+PQ+SQ=50 \\ \\ \\ (I) \ into \ (III): \\ \\ SQ+PQ+SQ=50 \\ \\ \therefore (IV) \ 2SQ+PQ=50 \\ \\ From \ (II): \\ \\ PQ=SQ-1 \\ \\ (II) \ into \ (IV): \\ \\ 2SQ+(SQ-1)=50 \\ 3SQ-1=50 \\ 3SQ=51 \\ \\ \boxed{SQ=17} \\ \\ \boxed{PS=17} \\ \\ PQ=SQ-1=17-1 \therefore \boxed{PQ=16}

Finally:

A=\sqrt{s(s-a)(s-b)(s-c)} \\ \\ A=\sqrt{s(s-PS)(s-SQ)(s-PQ)} \\ \\ A=\sqrt{s(s-17)(s-17)(s-16)} \\ \\ A=\sqrt{25(25-17)(25-17)(25-16)} \\ \\ \boxed{A=120}

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