# Please help solving this math question đ ?

Car A is driving at an average speed of 80km/hr at 6am.Car B is driving at an average speed of 100km/hr at 6.15am. What time will Car B catch up with Car A?

### 9 Answers

- KrishnamurthyLv 71 month agoFavourite answer
Car A is driving at an average speed of 80 km/hr at 6 am.

Car B is driving at an average speed of 100 km/hr at 6.15 am.Â

Car A is at a distance of 20 km when Car A begins to move.

Their relative speed is 100 - 80 = 20 km/hr

Car B catches up with Car A after 1 hour which is at 7:15 am

- 1 month ago
Assuming the 2 cars start from the same point

LetÂ Â t = 0Â Â be at 6:15 am

let d1 be the distance travelled by car A and write it as follows:

d1 = d0 + 80 km/hrÂ * tÂ Â Â Â

where d0 is the distance traveled by car A during the first 15 minutes between 6 am and 6:15 am at the speed of 80km/hrÂ

d0 = (6:15 - 6:00) * 80 km/hr = 15 minutes * 80 km/hrÂ

= 15 minutes * 80 km/(60 minutes) = 20 km

henceÂ

d1 = 20 km + 80 km/hrÂ *t

Let d2 be the distance travelled by car B starting at t = 0 (6:15 am) and write it as follows:

d2 = 100Â km/hrÂ * t

Car B catches up with Car A when d1 = d2

20km/hrÂ + 80km/hrÂ *t = 100km/hrÂ * t

Solve for tÂ

100km/hrÂ Â * t - 80km/hrÂ Â * t = 20 km

20 km/hrÂ * t = 20 km

t = 20 km / 20 km/hrÂ = 1 hrÂ

Car B catches up with car A 1 hour after t = 0 which corresponds to 6:15 am

Hence car B catches up with car A atÂ 6:15 am + 1 hrÂ = 7:15 am

more atÂ

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- PinkgreenLv 71 month ago
In 15/60 hr, A has travelled 80(15/60)=20 km.

Thus when B catches up A, B needs

15/60+20/(100-80)=1.25 hrs

=>

B catches up A at 7:15 am.

- DixonLv 71 month ago
Just to point out that if we only know the average speeds, we can't say where one will actually catch up with the other, we can only make the best guess.

- Ian HLv 71 month ago
At 6:15 A is ahead by 80 * (1/4) = 20 km

Their relative speed is 100 - 80 = 20 km/hr

Car B catches up with Car A after 1 hour which is at 7:15 am

- la consoleLv 71 month ago
Recall: s = d/t â where s is the speed, d is the distance, t is the time

The car A is driving at an average speed of 80 km/h at 6am.

The car B is driving at an average speed of 100 km/h at 6.15am.

So after 15 minutes, i.e. after (1/4) of an hour, the car A makes:

sâ = dâ/tâ

dâ = sâ.tâ

dâ = 80 * (1/4)

dâ = 20 km â this is the distance between the car A and the car B (after 15 minutes)

After 6.15am, the car A makes a distance:

sâ = dâ/t

dâ = sâ.t

After 6.15am, the car B makes a distance:

sâ = dâ/t

dâ = sâ.t

The car B will catch up with the car A when:

dâ = dâ + dâ

sâ.t = 20 + sâ.t

sâ.t - sâ.t = 20

t.(sâ - sâ) = 20

t = 20/(sâ - sâ)

t = 20/(100 - 80)

t = 20/(20)

t = 1 hour â this is the necessary time for the car B to cath up the car A

time = 6.15 + 1

time = 7.15am