Recall that speed is distance traveled over time:
v = d / t
Rearranging to get t,
t = d / v
For the two trips, the distance traveled is the same, while the speed and time of travel are different. We represent each time traveled as such:
Let t1 = time traveled (towards Rocky Mountain)
Let t2 = time traveled (return journey)
Thus,
t1 = d / v1
t1 = 135 / (2x^2 + 3x)
t2 = d / v2
t2 = 135 / (4x^2 - 9)
(a) The total flying time in hours:
t1 + t2 = [ 135 / (2x^2 + 3x) ] + [ 135 / (4x^2 - 9) ]
I'll leave the simplification of this expression to you.
(b) Substituting x=6 to the expression,
t1 + t2 = [ 135 / (2x^2 + 3x) ] + [ 135 / (4x^2 - 9) ]
t1 + t2 = [ 135 / (2(6)^2 + 3(6)) ] + [ 135 / (4(6)^2 - 9) ]
t1 + t2 = 135/90 + 135/135
t1 + t2 = 1.5 + 1
t1 + t2 = 2.5 hours
To check if the simplified expression you got in (a) is correct, try substituting x=6, and if the answer you got is the same as in (b), then it's correct.
Hope this helps~ `u`
1. A helicopter left Calgary and travelled 135 km west into the Rocky Mountains at an average speed of 2x^2 + 3x km/h. The return journey was at an average speed of 4x^2 - 9 km/h.
a) Write and simplify an expression for the total flying time in hours.
b) If the value of x is 6, determine the total flying time.
Can you please show your work/explain your answer? Thank you.
1 answer