# 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

Algebra ->  -> 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      Log On

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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 Answer by ankor@dixie-net.com(16525)   (Show Source): 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 : : 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 : Plot the graph of this equation, and find when the hovercraft catches up with the car ferry. : 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 : Did I make this understandable to you? ankor@att.net : 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)