*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**.*distance = rate x time*

**showing all of your work**, this is your next step...

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.

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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...

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 = x_{0} +_{ }v_{0}t + (1/2)at^{2}

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, (v_{car 1}) x t = (v_{car 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:

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