Question

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?

Answer 1

The time taken by Caleb is less than time taken by Alyssa , so, Caleb arrived earlier by 2 hours

Explanation:

Given in question as :

The Total distance which have to travel to both Alyssa and Caleb = 210 miles

Now, Starting distance travel by Alyssa (Da) = 105 miles

And The time taken to travel 105 miles (Ta)= 3.5 hours

Again , Starting distance travel by Caleb (Dc) = 168 miles

And The time taken to travel 168 miles (Tc)= 4 hours

SO, The speed at which Alyssa drove (Sa) =
`(Diatance)/(Time)`

Or, The speed at which Alyssa drove (Sa) =
`(Da)/(Ta)`

Or, The speed at which Alyssa drove (Sa) =
`(105 miles)/(3.5 hours)`

So, The speed at which Alyssa drove (Sa) = 30 miles per hour

Similarly for Caleb

SO, The speed at which Caleb drove (Sc) =
`(Diatance)/(Time)`

Or, The speed at which Caleb drove (Sc) =
`(Dc)/(Tc)`

Or, The speed at which Caleb drove (Sc) =
`(168 miles)/(4 hours)`

So, The speed at which Caleb drove (Sc) = 42 miles per hour

Now, to cover the distance of 210 miles

Time taken by Alyssa = Ta =
`(Da)/(Sa)`

Or, Ta =
`(210)/(30)` = 7 hours

∴ Time taken by Alyssa = 7 hours

Again ,

Now, to cover the distance of 210 miles

Time taken by Alyssa = Tc =
`(Dc)/(Sc)`

Or, Tc =
`(210)/(42)` = 5 hours

∴ Time taken by Caleb = 5 hours

`Time taken by Alyssa> Time taken by Caleb`

Or, Time taken by Alyssa - Time taken by Alyssa = 7 - 5 = 2 hours

Hence The time taken by Caleb is less than time taken by Alyssa , so, Caleb arrived earlier by 2 hours Answer