Please help solving this math question 😅?

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?

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  • 2 months ago
    Favourite 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. 

    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

  • 2 months ago

    Thanks for the help ;)

  • 2 months ago

    Assuming the 2 cars start from the same point

    Let   t = 0   be at 6:15 am

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

    d1 = d0 + 80 km/hr  * t      

    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/hr 

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

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

    hence 

    d1 = 20 km + 80 km/hr  *t

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

    d2 = 100  km/hr  * t

    Car B catches up with Car A when d1 = d2

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

    Solve for t 

    100km/hr   * t - 80km/hr   * t = 20 km

    20 km/hr  * t = 20 km

    t = 20 km / 20 km/hr  = 1 hr 

    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 at  6:15 am + 1 hr  = 7:15 am

    more at 

    https://www.analyzemath.com/math_problems/rate_tim...

  • 2 months ago

    ure muum ,,,,,,,,,,,

    Source(s): ur mum
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  • 2 months 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.

  • Dixon
    Lv 7
    2 months 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 H
    Lv 7
    2 months 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

  • 2 months ago

    Recall: s = d/t → 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:

    s₀ = d₀/t₀

    d₀ = s₀.t₀

    d₀ = 80 * (1/4)

    d₀ = 20 km ← this is the distance between the car A and the car B (after 15 minutes)

    After 6.15am, the car A makes a distance:

    s₁ = d₁/t

    d₁ = s₁.t

    After 6.15am, the car B makes a distance:

    s₂ = d₂/t

    d₂ = s₂.t

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

    d₂ = d₀ + d₁

    s₂.t = 20 + s₁.t

    s₂.t - s₁.t = 20

    t.(s₂ - s₁) = 20

    t = 20/(s₂ - s₁)

    t = 20/(100 - 80)

    t = 20/(20)

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

    time = 6.15 + 1

    time = 7.15am

  • Bryce
    Lv 7
    2 months ago

    80t + 20= 100t

    20t= 20; t= 1 hour

    7.15 am

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