If the average speed for a trip is reduced by 20%, by what percent does the time change?
let the distance be d
and the first rate be t
original time = d/t
new rate = .8t
new time = d/.8t = 1.25 d/t
so the time increased by 25%
To find the percent change in time, we need to understand the relationship between average speed, distance, and time.
The formula to calculate the time taken for a trip is:
Time = Distance / Speed
Let's assume the initial average speed for the trip is "S", and the initial time taken is "T".
So, initially, we have:
T = Distance / S
If the average speed is reduced by 20%, the new average speed would be 80% of the original speed, or 0.8S.
The new time taken for the trip, let's call it "T'", can be calculated using the same formula:
T' = Distance / (0.8S)
To find the percent change in time, we need to compare T' with T. The formula to calculate the percent change is:
Percent Change = ((New Value - Old Value) / Old Value ) * 100
In this case, the new value is T' and the old value is T.
So, calculating the percent change in time:
Percent Change = ((T' - T) / T) * 100
Substituting T' and T:
Percent Change = ((Distance / (0.8S)) - (Distance / S)) / (Distance / S) * 100
Simplifying the equation:
Percent Change = ((1/0.8) - 1) * 100
Percent Change = (1.25 - 1) * 100
Percent Change = 0.25 * 100
Percent Change = 25%
Therefore, reducing the average speed by 20% results in a 25% increase in the time taken for the trip.