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Calculus-particle motion

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1. A particle is moving on the x-axis (or any number line) Its position x(t), or distance from the origin, at the time t is given by x(t)=4t^3-16t^2+15t. t is greater than or equal to 0

a.) Where is the particle when it is at rest?

b.) During what time interval is the particle moving to the LEFT?

c.) When is the acceleration lest?

d.) What is the TOTAL distance traveled by the particle from t=0 to t=3 ?

  • Calculus-particle motion -

    a) at rest means dx/dt = 0
    12 t^2 -32 t + 15 = 0
    t = 2.05 or .607
    find x for those two values of t

    b) where is dx/dt negative
    do a sketch of dx(t)/dt =12t^2-32t+15
    you know it is a parabola facing up (holds water) because of the positive coef of x^2
    You know it is 0 at t = 2.05 and .607
    so it is negative between those two times.

    c) a = d^2x/dt^2 = 24 t -32
    that seems to be least for t = 0 unless it means |a| which is 0 at t = 32/24

    d) integral from 0 to 3 of original function is
    value of t^4 - (16/3)t^3 +(15/2)t^2
    when t = 3
    (since it is 0 at t = 0)

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