Arithmetic question with distance speed and time












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$begingroup$


If I have a time of departure with is 10:35 and I have an arrival time of 12:55 and the distance 36km how do I calculate average speed?
I get the formula of distance speed and time but it says the answer is 15 3/7 km/hr and I’m really confused on how it is.






arithmetic







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$endgroup$












  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Calvin Godfrey
    Jan 7 at 17:08






  • 1




    $begingroup$
    Avg speed = total distance / total time. Have you computed these?
    $endgroup$
    – Ben W
    Jan 7 at 17:08










  • $begingroup$
    What is the elapsed time in hours from $10$:$35$ to $12$:$55$?
    $endgroup$
    – KM101
    Jan 7 at 17:09












  • $begingroup$
    Tried finding the differences of the times which got me 02:20 and I had to covert it to 2 1/3 hours and then I know I have to divide 36 by something but that’s as far as I got really
    $endgroup$
    – baek won
    Jan 7 at 17:10






  • 1




    $begingroup$
    $2 frac{1}{3} = frac{7}{3}$ and $frac{36}{frac{7}{3}} = frac{36cdot 3}{7} = frac{108}{7} = 15frac{3}{7}$.
    $endgroup$
    – KM101
    Jan 7 at 17:13
















0












$begingroup$


If I have a time of departure with is 10:35 and I have an arrival time of 12:55 and the distance 36km how do I calculate average speed?
I get the formula of distance speed and time but it says the answer is 15 3/7 km/hr and I’m really confused on how it is.






arithmetic







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Calvin Godfrey
    Jan 7 at 17:08






  • 1




    $begingroup$
    Avg speed = total distance / total time. Have you computed these?
    $endgroup$
    – Ben W
    Jan 7 at 17:08










  • $begingroup$
    What is the elapsed time in hours from $10$:$35$ to $12$:$55$?
    $endgroup$
    – KM101
    Jan 7 at 17:09












  • $begingroup$
    Tried finding the differences of the times which got me 02:20 and I had to covert it to 2 1/3 hours and then I know I have to divide 36 by something but that’s as far as I got really
    $endgroup$
    – baek won
    Jan 7 at 17:10






  • 1




    $begingroup$
    $2 frac{1}{3} = frac{7}{3}$ and $frac{36}{frac{7}{3}} = frac{36cdot 3}{7} = frac{108}{7} = 15frac{3}{7}$.
    $endgroup$
    – KM101
    Jan 7 at 17:13














0












0








0





$begingroup$


If I have a time of departure with is 10:35 and I have an arrival time of 12:55 and the distance 36km how do I calculate average speed?
I get the formula of distance speed and time but it says the answer is 15 3/7 km/hr and I’m really confused on how it is.






arithmetic







share|cite|improve this question











$endgroup$




If I have a time of departure with is 10:35 and I have an arrival time of 12:55 and the distance 36km how do I calculate average speed?
I get the formula of distance speed and time but it says the answer is 15 3/7 km/hr and I’m really confused on how it is.






arithmetic




arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 17:21









David G. Stork

10.7k31332




10.7k31332










asked Jan 7 at 17:07









baek wonbaek won

11




11












  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Calvin Godfrey
    Jan 7 at 17:08






  • 1




    $begingroup$
    Avg speed = total distance / total time. Have you computed these?
    $endgroup$
    – Ben W
    Jan 7 at 17:08










  • $begingroup$
    What is the elapsed time in hours from $10$:$35$ to $12$:$55$?
    $endgroup$
    – KM101
    Jan 7 at 17:09












  • $begingroup$
    Tried finding the differences of the times which got me 02:20 and I had to covert it to 2 1/3 hours and then I know I have to divide 36 by something but that’s as far as I got really
    $endgroup$
    – baek won
    Jan 7 at 17:10






  • 1




    $begingroup$
    $2 frac{1}{3} = frac{7}{3}$ and $frac{36}{frac{7}{3}} = frac{36cdot 3}{7} = frac{108}{7} = 15frac{3}{7}$.
    $endgroup$
    – KM101
    Jan 7 at 17:13


















  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Calvin Godfrey
    Jan 7 at 17:08






  • 1




    $begingroup$
    Avg speed = total distance / total time. Have you computed these?
    $endgroup$
    – Ben W
    Jan 7 at 17:08










  • $begingroup$
    What is the elapsed time in hours from $10$:$35$ to $12$:$55$?
    $endgroup$
    – KM101
    Jan 7 at 17:09












  • $begingroup$
    Tried finding the differences of the times which got me 02:20 and I had to covert it to 2 1/3 hours and then I know I have to divide 36 by something but that’s as far as I got really
    $endgroup$
    – baek won
    Jan 7 at 17:10






  • 1




    $begingroup$
    $2 frac{1}{3} = frac{7}{3}$ and $frac{36}{frac{7}{3}} = frac{36cdot 3}{7} = frac{108}{7} = 15frac{3}{7}$.
    $endgroup$
    – KM101
    Jan 7 at 17:13
















$begingroup$
What have you tried so far?
$endgroup$
– Calvin Godfrey
Jan 7 at 17:08




$begingroup$
What have you tried so far?
$endgroup$
– Calvin Godfrey
Jan 7 at 17:08




1




1




$begingroup$
Avg speed = total distance / total time. Have you computed these?
$endgroup$
– Ben W
Jan 7 at 17:08




$begingroup$
Avg speed = total distance / total time. Have you computed these?
$endgroup$
– Ben W
Jan 7 at 17:08












$begingroup$
What is the elapsed time in hours from $10$:$35$ to $12$:$55$?
$endgroup$
– KM101
Jan 7 at 17:09






$begingroup$
What is the elapsed time in hours from $10$:$35$ to $12$:$55$?
$endgroup$
– KM101
Jan 7 at 17:09














$begingroup$
Tried finding the differences of the times which got me 02:20 and I had to covert it to 2 1/3 hours and then I know I have to divide 36 by something but that’s as far as I got really
$endgroup$
– baek won
Jan 7 at 17:10




$begingroup$
Tried finding the differences of the times which got me 02:20 and I had to covert it to 2 1/3 hours and then I know I have to divide 36 by something but that’s as far as I got really
$endgroup$
– baek won
Jan 7 at 17:10




1




1




$begingroup$
$2 frac{1}{3} = frac{7}{3}$ and $frac{36}{frac{7}{3}} = frac{36cdot 3}{7} = frac{108}{7} = 15frac{3}{7}$.
$endgroup$
– KM101
Jan 7 at 17:13




$begingroup$
$2 frac{1}{3} = frac{7}{3}$ and $frac{36}{frac{7}{3}} = frac{36cdot 3}{7} = frac{108}{7} = 15frac{3}{7}$.
$endgroup$
– KM101
Jan 7 at 17:13










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$begingroup$

Once you calculated 2 1/3 hours, you can figure out what to do next with units. The answer is speed, which has units (distance)/(time). So you divide the distance (36 km) by the time (2 1/3 hours) to get 15 3/7 km/hr.






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    $begingroup$

    Once you calculated 2 1/3 hours, you can figure out what to do next with units. The answer is speed, which has units (distance)/(time). So you divide the distance (36 km) by the time (2 1/3 hours) to get 15 3/7 km/hr.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Once you calculated 2 1/3 hours, you can figure out what to do next with units. The answer is speed, which has units (distance)/(time). So you divide the distance (36 km) by the time (2 1/3 hours) to get 15 3/7 km/hr.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Once you calculated 2 1/3 hours, you can figure out what to do next with units. The answer is speed, which has units (distance)/(time). So you divide the distance (36 km) by the time (2 1/3 hours) to get 15 3/7 km/hr.






        share|cite|improve this answer









        $endgroup$



        Once you calculated 2 1/3 hours, you can figure out what to do next with units. The answer is speed, which has units (distance)/(time). So you divide the distance (36 km) by the time (2 1/3 hours) to get 15 3/7 km/hr.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 17:13









        Calvin GodfreyCalvin Godfrey

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