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Math Help - who I can find the displacement in this Q ?

  1. #1
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    who I can find the displacement in this Q ?

    Hi all

    who I can find the displacement in this Q ?

    A jetliner lands with a speed of 90 m/s on runway . It runs with constant speed for 2 s , before the brakes are applied . The brakes produce a deceleration of 20 m/s2 . What is its displacement on the runway .

    I try to solve

    V0 = 90 m/s
    t1 = 2s
    x = V0 X t1 = 180 m
    a = -20 m/s2

    then I don't know who I complte it ?
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  2. #2
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    maybe u should use this formula
    s=1/2(v+u)t

    s is displacement
    v is initial velocity
    u is final velocity
    t is time

    s=1/2(90+70)2
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  3. #3
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    Quote Originally Posted by r-soy View Post
    Hi all

    who I can find the displacement in this Q ?

    A jetliner lands with a speed of 90 m/s on runway . It runs with constant speed for 2 s , before the brakes are applied . The brakes produce a deceleration of 20 m/s2 . What is its displacement on the runway .

    I try to solve

    V0 = 90 m/s
    t1 = 2s
    x = V0 X t1 = 180 m
    a = -20 m/s2

    then I don't know who I complte it ?
    180 + x where x is calculated below:

    initial velocity = 90 m/s.
    final velocity = 0 m/s.
    a = 20 m/s^2.
    x = ?

    Now use the usual formula.
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  4. #4
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    Hi,

    You could try drawing a rough graph, then calculate the area under the graph for the total distance

    It starts at 90m/s, so in a speed time graph there is a horizontal line at 90 for 2 sec. Then the line slopes downwards, losing 20m's every second, so in 4.5 seconds it has a speed of 0.

    When you calculate the area under the graph:

    2x90 + (6.5-2)x90x0.5 = 382.5m
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  5. #5
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    Quote Originally Posted by Illusion3 View Post
    Hi,

    You could try drawing a rough graph, then calculate the area under the graph for the total distance

    It starts at 90m/s, so in a speed time graph there is a horizontal line at 90 for 2 sec. Then the line slopes downwards, losing 20m's every second, so in 4.5 seconds it has a speed of 0.

    When you calculate the area under the graph:

    2x90 + (6.5-2)x90x0.5 = 382.5m
    hI

    which formula you used here ??

    and can help me in the gragh why you say in 4.5 seconds it has a speed of 0.
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  6. #6
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    Quote Originally Posted by r-soy View Post
    hI

    which formula you used here ??

    and can help me in the gragh why you say in 4.5 seconds it has a speed of 0.
    Reply #3 has given you a very simple approach. Why are you trying to do it this way?
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  7. #7
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    Quote Originally Posted by r-soy View Post
    hI

    which formula you used here ??

    and can help me in the gragh why you say in 4.5 seconds it has a speed of 0.
    There is a general rule that the area under a speed time graph is the distance traveled. So off course you can just calculate the area under it just like calculating the area of a square and triangle.

    Because the deceleration is 20 meters per second per second, this means that a speed of 20meters per second is lost every second. So it takes 4.5 seconds for a speed of 90m/s to reach 0m/s (90/20 = 4.5)
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