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anonymous

  • 4 years ago

A particle moves along a straight line and its position at time t is given by s(t)=2t3−15t2+24t where s is measured in feet and t in seconds. a) Find the velocity (in ft/sec) of the particle at time t=0 b) The particle stops moving twice, once when t=A and again when t=B where A<B what are A and B c) What is the position of the particle at time 10? d) The TOTAL distance the particle travels between time 0 and time 10 is

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  1. anonymous
    • 4 years ago
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    \[v(t)=s'(t)=6t^2-30t+24\] \[v(0)=24 ft/s\] b) If it stops moving, the velocity is 0: \[6t^2-30t+24=0, t^2-5t+4=0, t=1, t=4\]a)\[v(t)=s'(t)=6t^2-30t+24\] c)\[s(10)=2(10)^3−15(10)^2+24(10)=740 ft \]

  2. anonymous
    • 4 years ago
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    how do you figure out the total distance

  3. anonymous
    • 4 years ago
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    also can you help me with this Find equations of the tangent line and normal line to the curve y=10cosx at the point (π/3,5). Derivative of y= -10sinx The slope of the tangent line is m1=-10sin(pi/3) The equation of the tangent line is y=-5sqrt(3)x+(5pi/sqrt3)+5 The slope of the normal line is m2=? The equation of the normal line is y=? I just need to know the last two

  4. anonymous
    • 4 years ago
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    the particle changes direction at t = 1 and t = 4 displacement at t = 1 = 11 (plug 1 into formula for s(t) at time = 4 displacement is -16 so it hase travelled 11 + 11 + 16 = 38 up to then then another 16 + 740 by t=10 total distance travelled = 38 + 16 + 740 = 794 ft

  5. anonymous
    • 4 years ago
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    ahh jimmy thanks very much. can you take a look at the question above it too please

  6. anonymous
    • 4 years ago
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    not above but below

  7. anonymous
    • 4 years ago
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    the one about normal and tangent line

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