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Math Help - Meeting point between two vehicles

  1. #1
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    Meeting point between two vehicles

    Driver A starts his car and accelerates with a constant accelerasion of 0.8 m/s^2.
    Driver B starts and follows on a motorcycle from the same spot, 6,0 sec later.
    Driver B has got a constant acceleration of 1.8 m/s^2. (Neither of them accelerate higher than 80 km/h)

    a) How long does it take driver B to catch up with driver A?
    b) How long have they driven when they meet?
    c) How fast do each driver drive when they meet?

    I have tried the four formulas:

    v = at+v0

    s = 1/2 (v+v0) * t

    s = 1/2 a * t^2 + v0 * t

    2 * a * s = v^2 - v0^2

    With no luck at all. I am confident that some calculations might be wrong.

    All help is appreciated, hopefully a quick response.
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  2. #2
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    Re: Meeting point between two vehicles

    Quote Originally Posted by roahau View Post
    Driver A starts his car and accelerates with a constant accelerasion of 0.8 m/s^2.
    Driver B starts and follows on a motorcycle from the same spot, 6,0 sec later.
    Driver B has got a constant acceleration of 1.8 m/s^2. (Neither of them accelerate higher than 80 km/h)

    a) How long does it take driver B to catch up with driver A?
    b) How long have they driven when they meet?
    c) How fast do each driver drive when they meet?
    for driver A ...

    v_A = 0.8t

    s_A = 0.4t^2

    for driver B ...

    v_B = 1.8(t-6)

    s_B = 0.9(t-6)^2

    for (a) and (b) , set s_A = s_B , solve for t to find the time driver A's run ... don't forget that driver B's run is (t-6). once you get the time, find the value of s_A and s_B (which should be the same, right?)

    for (c) , determine the values of v_A and v_B at the time found above.
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  3. #3
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    Re: Meeting point between two vehicles

    Quote Originally Posted by skeeter View Post
    for driver A ...

    v_A = 0.8t

    s_A = 0.4t^2

    for driver B ...

    v_B = 1.8(t-6)

    s_B = 0.9(t-6)^2

    for (a) and (b) , set s_A = s_B , solve for t to find the time driver A's run ... don't forget that driver B's run is (t-6). once you get the time, find the value of s_A and s_B (which should be the same, right?)

    for (c) , determine the values of v_A and v_B at the time found above.
    How am I supposed to determine t without the values of vA and vB ? I can't use the first formula without time, and I can't use the time forumula without the final velocity
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  4. #4
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    Re: Meeting point between two vehicles

    Quote Originally Posted by roahau View Post
    How am I supposed to determine t without the values of vA and vB ? I can't use the first formula without time, and I can't use the time forumula without the final velocity
    when the two vehicles meet, they are at the same position.

    s_A = s_B

    0.4t^2 = 0.9(t-6)^2

    solve for t , then proceed as I stated in my previous post.
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