SOLUTION: a-A cross- channel hovercraft ferry leaves Dover harbour at 9am travelling at 35km/hour. Find an equation giving d (the distance travelled in km) in terms of t (the time in hours a
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-> SOLUTION: a-A cross- channel hovercraft ferry leaves Dover harbour at 9am travelling at 35km/hour. Find an equation giving d (the distance travelled in km) in terms of t (the time in hours a
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Question 215948This question is from textbook Longman mathematics for IGCSE book1
: a-A cross- channel hovercraft ferry leaves Dover harbour at 9am travelling at 35km/hour. Find an equation giving d (the distance travelled in km) in terms of t (the time in hours after 9am). Plot the graph of this equatiopn for 0
b-As the hovercraft leaves the harbour, the captain sees a car ferry 5 km ahead travelling in the same direction at 15 km/hour. Find an equation for d in terms of t for the car ferry. Plot the graph of this equation, and find when the hovercraft catches up with the car ferry.
Please answer both of the parts. I am having a serious problem in this question.
Thankyou ver very much indeed. This question is from textbook Longman mathematics for IGCSE book1
You can put this solution on YOUR website! a-A cross- channel hovercraft ferry leaves Dover harbour at 9am travelling at 35km/hour.
Find an equation giving d (the distance traveled in km) in terms of t (the time in hours after 9am).
d = 35t
Plot the graph of this equation for 0<1.
for t = 1 hr; d = 35
for t = 2 hr; d = 70
Plot two points; x=1; y=35; and x=2; y=70
you can label your graph as 0 is 9AM; 1 is 10Am, etc
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b-As the hovercraft leaves the harbour, the captain sees a car ferry 5 km ahead traveling in the same direction at 15 km/hour. Find an equation for d in terms of t for the car ferry.
Since it has a 5 km head start, the equation is d = 15t + 5
Plot this one the same way: t = 1; d=20; t = 2; d = 35
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Plot the graph of this equation, and find when the hovercraft catches up with the car ferry.
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You can solve the two equations, when they are both the same distance
35t = 15t + 5
35t - 15t = 5
20t = 5
t =
t = hr which is 15 min to catch up, agrees with the graph
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Did I make this understandable to you? ankor@att.net
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To find the dist, substitute for t in either equation
we can use .25 hr
d = 35t
d = 35 * .25
d = 8.75 km
or
d = 15t + 5
d = 15(.25) + 5
d = 3.75 + 5
d = 8.75 km
;
Which you would expect when both vessels are at the same point in 15 min
Change the scale in the graph to see it (the +x axis represents 1 hr)
