At 9:00 am, a car leaves a certain tollbooth heading north, At 9:15 am, a

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Star27

New member
car trveling 10 km/hr faster than? At 9:00 am, a car leaves a certain tollbooth heading north, At 9:15 am, a car trveling 10 km/hr faster than the first car leaves the same tollbooth heading north. If the second car overtaes the first car at 11:15 m, find the speed of the slower car
 
speed = distance/time
distance = speed x time
When the second car overtakes the first car, the distance will be the same for both of them.
let s = speed of slower car
s(11:15-9:00 hr) = (s + 10km/hr)(11:15-9:15)
s2.25hr = (s + 10km/hr)(2hr)
s2.25hr = s2hr +10km/hr(2hr)
2.25s-2s = 5km/hr
0.25s = 5km/hr
s = 20km/hr
 
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