Question

Joseph is driving at an average of 50 miles per hour on a 330-mile road trip. On the way back, Joseph was about to average 60 miles per hour. How much faster was the trip back home

Answer 1

1 second faster was the trip back home .

Step-by-step explanation:

Here we have , Joseph is driving at an average of 50 miles per hour on a 330-mile road trip. On the way back, Joseph was about to average 60 miles per hour.  We need to find  How much faster was the trip back home . Let's find out:

• Joseph is driving at an average of 50 miles per hour on a 330-mile road trip.

We know that , $Speed = \frac{Distance}{time}$

⇒ $Speed = \frac{Distance}{time}$

⇒ $50 = \frac{330}{time}$

⇒ $time = \frac{330}{50}$

⇒ $time = 6.6.sec$

• On the way back, Joseph was about to average 60 miles per hour.  

We know that , $Speed = \frac{Distance}{time}$

⇒ $Speed = \frac{Distance}{time}$

⇒ $60 = \frac{330}{time}$

⇒ $time = \frac{330}{60}$

⇒ $time = 5.5sec$

So, Time difference in both trips is $6.5sec-5.5sec=1sec$ .

Therefore, 1 second faster was the trip back home .