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    calc problem rectilineal motion

    How would you use the anitderivative to solve this problem?

    A car is traveling at 50mi/hr when the brakes are fully applied, producing a constant deceleration of 22ft/sec^2. What is the distance covered before the car comes to a stop? Thanks.
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    Quote Originally Posted by UMStudent View Post
    How would you use the anitderivative to solve this problem?

    A car is traveling at 50mi/hr when the brakes are fully applied, producing a constant deceleration of 22ft/sec^2. What is the distance covered before the car comes to a stop? Thanks.
    The general equation which describes this motion is:

    s(t) = -(1/2)at^2+v_0(t)+s_0

    Where,
    s(t) is distance traveled from starting point.
    "a" is the acceleration.
    v_0 is initial velocity.
    s_0 is initial starting distance traveled (which is zero here).

    Thus,
    s(t) = - (1/2)*11*t^2+50*t

    s(t) = -11*t^2+50t

    Use this to solve all your problems.
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    Quote Originally Posted by UMStudent View Post
    How would you use the anitderivative to solve this problem?

    A car is traveling at 50mi/hr when the brakes are fully applied, producing a constant deceleration of 22ft/sec^2. What is the distance covered before the car comes to a stop? Thanks.
    I am going to use a coordinate system where the origin is at the point where deceleration started and +x is the direction of the initial velocity.

    a = -22 ft/s^2 and Initial speed = 50 mi/hr = 73.33 ft/s
    (a is negative since it is opposing the direction of the initial velocity.)

    Now:
    v(t) = Int[a(t'), 0, t] + v0 <-- Where v0 is an integration constant.

    v(t) = Int[-22, 0, t] + v0 = -22t + v0

    Now, at t = 0 s, v(0) = 73.33 ft/s, so
    v(t) = -22t + 73.33 ft/s

    Again:
    x(t) = Int[v(t'), 0, t] + x0 <-- Where x0 is an integration constant.

    x(t) = Int[-22t' + 73.33, 0, t]

    x(t) = (1/2)(-22)t^2 + 73.33t + x0

    At t = 0 s, x(0) = 0 ft, so
    x(t) = -11t^2 + 73.33t

    When does the car come to a stop? When v(t) = 0 ft/s. Thus when:
    v(t) = -22t + 73.33 = 0

    t = 73.33/22 s = 3.333 s

    So where is the car at this time?
    x(3.333) = -11(3.333)^2 + 73.33(3.333) = 122.22 ft.

    Thus the car has travelled:
    d = x(3.333) - x(0) = 122.22 ft - 0 ft = 122.22 ft

    -Dan
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