Question

Richard is driving home to visit his parents. 135 mi of the trip are on the interstate highway where the speed limit is 65 mph . Normally Richard drives at the speed limit, but today he is running late and decides to take his chances by driving at 73 mph .

Answer 1

Richard will save approximately 13.68 minutes by driving at 73 mph for 135 miles instead of at the speed limit of 65 mph. The mathematical concept used involves calculating travel time at different speeds and finding the difference between these times.

Step-by-step explanation:

The student's question involves calculating how much time Richard will save by driving at 73 mph instead of the speed limit of 65 mph for a distance of 135 miles. This is a mathematics problem that falls under the category of rate, time, and distance, which is often covered in high school math classes.

To solve this, we need to calculate the time taken to travel 135 miles at both speeds and then find the difference in time. At 65 mph, the time taken (T1) is 135 miles / 65 mph. At 73 mph, the time taken (T2) is 135 miles / 73 mph. The time saved is T1 - T2.

Using the formula time = distance / speed, we get:

Time at 65 mph: T1 = 135 mi / 65 mph = 2.077 hours

Time at 73 mph: T2 = 135 mi / 73 mph = 1.849 hours

The time saved is T1 - T2 = 2.077 hours - 1.849 hours = 0.228 hours, which is about 13.68 minutes.

Answer 2

He saves 13.2 minutes

Step-by-step explanation:

Hey! The question is incomplete, but it can be found on the internet. The question is:

How many minutes did he save?

Let's call:

`t_(1):Time \ at \ speed \ 65mph \\ \\ t_(2):Time \ at \ speed \ 73mph \\ \\ v_(1)=65mph \\ \\ v_(2)=73mph`

We know that the 135 miles are on the interstate highway where the speed limit is 65 mph. From this, we can calculate the time it takes to drive on this highway. Assuming Richard maintains constant the speed:

`v=(d)/(t) \\ \\ d:distance \\ \\ t:time \\ \\ v:velocity \\ \\ t_(1)=(d)/(v_(1)) \\ \\ t=(135)/(65) \\ \\ t_(1)=2.07 \ hours`

Today he is running late and decides to take his chances by driving at 73 mph, so the new time it takes to take the trip is:

`t_(2)=(135)/(73) \\ \\ t_(2)=1.85 \ hours`

So he saves the time
`t_(s)`:

`t_(s)=t_(1)-t_(2)=2.07-1.85=0.22 \ hours`

In minutes:

`t_(s)=0.22h\left((60min)/(1h)\right) \\ \\ \boxed{t_(s)=13.2min}`