Q:

Alyssa and Caleb both drove 210 miles to the beach in separate cars. They left at the same time. They both drove at constant speed. Alyssa drove 105 miles in 3.5 hours. Caleb drove 168 miles is 4 hours. Who arrived at earlier? How much earlier?

Accepted Solution

A:
Answer:The time taken by Caleb is less than time taken by Alyssa , so, Caleb arrived earlier by 2 hoursStep-by-step explanation:Given in question as :The Total distance which have to travel to both Alyssa and Caleb = 210 milesNow, Starting distance travel by Alyssa (Da) = 105 milesAnd    The time taken to travel 105 miles (Ta)= 3.5 hoursAgain , Starting distance travel by Caleb (Dc) = 168 milesAnd    The time taken to travel 168 miles (Tc)= 4 hours SO, The speed at which Alyssa drove (Sa)  = [tex]\frac{Diatance}{Time}[/tex]Or, The speed at which Alyssa drove (Sa)  = [tex]\frac{Da}{Ta}[/tex]Or, The speed at which Alyssa drove (Sa)  = [tex]\frac{105 miles}{3.5 hours}[/tex]So, The speed at which Alyssa drove (Sa)  = 30 miles per hourSimilarly for CalebSO, The speed at which Caleb drove (Sc)  = [tex]\frac{Diatance}{Time}[/tex]Or, The speed at which Caleb drove (Sc)  = [tex]\frac{Dc}{Tc}[/tex]Or, The speed at which Caleb drove (Sc)  = [tex]\frac{168 miles}{4 hours}[/tex]So, The speed at which Caleb drove (Sc)  = 42 miles per hourNow, to cover the distance of 210 milesTime taken by Alyssa = Ta =  [tex]\frac{Da}{Sa}[/tex]Or,                                   Ta = [tex]\frac{210}{30}[/tex] = 7 hours∴ Time taken by Alyssa = 7 hoursAgain ,Now, to cover the distance of 210 milesTime taken by Alyssa = Tc =  [tex]\frac{Dc}{Sc}[/tex]Or,                                   Tc = [tex]\frac{210}{42}[/tex] = 5 hours∴ Time taken by Caleb = 5 hours[tex]Time taken by Alyssa> Time taken by Caleb[/tex]Or, Time taken by Alyssa - Time taken by Alyssa  = 7 - 5 = 2 hoursHence The time taken by Caleb is less than time taken by Alyssa , so, Caleb arrived earlier by 2 hours      Answer