Drivers a and b travel west from boston. driver a leaves three hours earlier traveling at a constant speed of 68mph. driver b follows at a constant speed of 85 mph. how many hours would it take for driver b to catch up to driver a?

We can answer this using one of the equations of linear motion:

v = d / t

where:

v = velocity

d = distance

t = time

In the problem, we are asked to find for the time in which Driver B will catch up to Driver A. Therefore,  find the time when dA = dB. Rearranging the equation and equation dA and dB will result in:

vA * tA = vB * tB  ---> 1

It was given that:

vA = 68 mph

tA = tB + 3 (since person A was travelling 3 hours earlier)

vB = 85 mph

tB = unknown

Substituting into equation 1:

68 * (tB + 3) = 85 * tB

68 tB + 204 = 85 tB

tB = 12 hrs

Therefore driver B would catch up to driver A after 12 hrs.

he made 11 calls in one day