The instrument reported back that the number of cars passing each minute varied linearly between 6.30am and 8.30am. The instrument reported 55 cars passing in 1 minute at 7.00am while at 8.00am, 115 cars passed in 1 minute. Fully communicate your working to do the following.
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First work out the gradient. If t represents the time in minutes after 6.30am and c the number of cars passing per minute, then the gradient is (difference in numbers of cars)/(difference in time)=(115-55)/(90-30) because t=30 at 7.00am and t=90 at 8.00am. So the gradient is 60/60=1. We can now write c=t+A where we need to find the constant A. When t=30, c=55 so 55=30+A, making A=25. The linear equation is c=t+25.

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