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.
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`
To calculate the total flying time, we need to find the time taken for each leg of the journey and sum them together.
a) First, let's calculate the time taken for the outbound journey. The formula for time is distance divided by speed. In this case, the distance is 135 km and the speed is given by the expression 2x^2 + 3x km/h. So, the time taken for the outbound journey is:
Time_outbound = Distance / Speed_outbound
= 135 km / (2x^2 + 3x) km/h
Next, let's calculate the time taken for the return journey. The distance for the return journey is also 135 km, and the speed is given by the expression 4x^2 - 9 km/h. So, the time taken for the return journey is:
Time_return = Distance / Speed_return
= 135 km / (4x^2 - 9) km/h
Finally, we can find the total flying time by adding the time for the outbound and return journeys:
Total_flying_time = Time_outbound + Time_return
= 135 km / (2x^2 + 3x) km/h + 135 km / (4x^2 - 9) km/h
b) To determine the total flying time when x = 6, we substitute x = 6 into the expression for the total flying time:
Total_flying_time = 135 km / (2(6)^2 + 3(6)) km/h + 135 km / (4(6)^2 - 9) km/h
Now we can calculate the total flying time by simplifying the expressions and performing the calculations:
Total_flying_time = 135 km / (72 + 18) km/h + 135 km / (144 - 9) km/h
= 135 km / 90 km/h + 135 km / 135 km/h
= 3/2 h + 1 h
= 2 1/2 h
= 2.5 hours
Therefore, when x = 6, the total flying time is 2.5 hours.
a) To find the total flying time, we need to find the time taken for the outbound journey and the time taken for the return journey, and then add them together.
Let's start by finding the time taken for the outbound journey. The distance travelled is 135 km, and the average speed is given by the expression 2x^2 + 3x km/h. We can use the formula:
Time = Distance / Speed
So the time taken for the outbound journey is:
Time_outbound = 135 / (2x^2 + 3x)
Next, let's find the time taken for the return journey. The distance travelled is still 135 km, but the average speed is given by the expression 4x^2 - 9 km/h. Using the same formula as before:
Time_return = 135 / (4x^2 - 9)
To find the total flying time, we need to add the time for the outbound journey and the time for the return journey:
Total_flying_time = Time_outbound + Time_return
b) To determine the total flying time when x = 6, substitute x = 6 into the expressions for Time_outbound and Time_return, and then add them together.
Time_outbound = 135 / (2(6)^2 + 3(6))
= 135 / (2(36) + 18)
= 135 / (72 + 18)
= 135 / 90
= 1.5 hours
Time_return = 135 / (4(6)^2 - 9)
= 135 / (4(36) - 9)
= 135 / (144 - 9)
= 135 / 135
= 1 hour
Total_flying_time = Time_outbound + Time_return
= 1.5 + 1
= 2.5 hours
So, when x = 6, the total flying time is 2.5 hours.