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When does 2nd car catch 1st car?

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6 Answers

Hi Kayla;
All of the above answers are great.  However, when I took physics, this is the way we had to show all work.
The question is...
1st Car leaves point A at 5pm at avg speed of 80km/hr. Second car leaves point A 45 minutes later at 110km/hr. When does the 2nd car catch the 1st car. Show working.
Emphasis supplied
As Justin explains, distance = rate x time
The question is when does the distance of 1st car=2nd car
(rate)(time)=(rate)(time)
x=time in hours of first car.  (Note the specification of hours.)
(80 km/hr)(x)=(110 km/hr)(x-0.75hrs)
As part of showing all of your work, this is your next step...
(80 km/hr)(x)=(110 km/hr)(x-0.75)
The kilometers and hours cancel.  You need to cancel words as much as you need to cancel numbers.
80x=110(x -0.75)
We must isolate x.
Let's spread-out...
80x=110x-82.5
Let's divide both sides by 80...
80x/80=(110x-82.5)/80
x=1.375x-1.03125
Let's subtract 1.375x from both sides...
x-1.375x=-1.375x-1.03125
-0.375x=-1.03125
Let's divide both sides by -0.375
(-0.375x)/(-0.375) = -1.03125/-0.375
x= 2.75
After 2.75 hours from the time of departure of the first car, the second car caught-up with the first car.
According to Kirill, this is 7:45 pm.
 
 
distance = rate x time
 
now when the car "catches up" the distances that they drove will be equal
 
so that means the rate x time for each car has to come out to be the same (bc rate x time is distance as shown above)
 
so car 1 has a rate of 80km/k
 
so do 80T where T is the time it drove
 
car2 has a rate of 110km/h and it left 45 min after, so it was on the road 3/4 of an hour less than the other car
so do 110(T-.75) for car 2
 
 
now set them equal to each other and solve for T bc that will be the time (in hours) that it will take
 
80T = 110(T-.75)
Assuming that you are taking physics, another way you can do this is with kinematics (which is a bit longer but great practice for any tougher questions): 
 
We use the equation x = x0 + v0t + (1/2)at2
 
The initial position is taken as 0, and there is no acceleration since both cars are traveling at a constant avg speed, so we boil the equation down to x = v x t
 
When car 2 catches up with car 1, they both travel the same distance. Therefore, (vcar 1) x t = (vcar 2) x t. 
 
Car 2 starts 45 min after car 1, so car 2 is delayed by .75 hours. So, we need to account for this. 
 
(v car 1) x t = (v car 2) x (t- 0.75)
 
80t = 110 (t-0.75)
80t = 110t - 82.5
-30t = -82.5
t = 2.75 hours
 
Car 1 travels for 2.75 hours before Car 2 catches up. Therefore, the time it takes for Car 2 to reach Car 1 is 2.75 hours - 0.75 hours = 2 hours. 
 
 
Let t be the total hours the first car drives.  t-3/4 is the total time for the second car. (45 minutes is 3/4 hour)
The total distance of the first car from point A is 80*t,
the total distance of the second car from point A is 110*(t-3/4)
The distances are equal when 80t=110t-110*3/4 or when 80t=110t-82.5
80t+82.5=110t and so 30t=82.5 or t=2.75
For t=2.75 the first car drives 80*2.75=220 km
For t=2.75-.75 the second car drives 2*110=220km. 2.75 after 5 is 7:45
SECOND solution
Let the clock start at the second car start.  The first car is 80*3/4=60km away.
The total distance of the first car is 60+80t
The total distance of the second car is 110t
These distances are equal when 80t+60=110t or when 60=30t or t=2
At that time te first car is 60+2#80=60+160=220km away.
The second car is 110*2=220 km away.
2 hours after 5:45 is 7:45
1) In 45 min. the first car is at d=80 km/h*3/4 h=60 km from point A.
 
2) Now, imagine yourself sitting in the first car. For you, the second car comes at you with the speed s=110-80=30 km /h. The distance between you is 60 km, found in part 1). Therefore, the time it takes for the second car to catch you is:
t=60/30=2 h. Since the second car started at 5:45 p.m., it will catch the first car at (5:45)+2=7:45 p.m.
When the second car starts at 5:45pm, the first car will have driven a distance of 80 km/hr*0.75 hr=60 km. Since their relative speed is (110-80) km/h=30 km/h, it will take the second car 60 km/(30 km/h) = 2 h to catch up, which happens at 7:45pm.