Physics: You leave on a trip in order to attend a meeting that will start after you begin your trip. Along the way you plan to stop for dinner. If the fastest you can safely drive is 65 mi/h, what is the longest time you can spend over dinner and still arrive just in time for the meeting

You leave on a trip in order to attend a meeting - id5714281, annabellthedivp3u9ve

newyorkvincetti

22.01.2022

2.96 h

Explanation:

We are given that

Speed of man,v=65mi/h

Time ,t=9.9 h

Total distance,s=451 miles

We have to find the time which you can spend over dinner and still arrive juts in time for the meeting.

Distance, $d=v\times t$

Using the formula

$d=65\times 9.9=643.5 miles$

Difference between distance=643.5-451=192.5 miles

Time,= $t'=\frac{difference\;in\;distance}{speed}$

$t'=\frac{192.5}{65}=2.96 h$

Hence, you can spend maximum time over dinner=2.96 h

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tlturner03

01.04.2020

Answer is in the photo. I can only upload it to a file hosting service. link below!

is.gd/qKzpBU

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