Mathematics

Question

find the body's position at time t a=10cos 5t, v (0)= -10,s (0)=10

1 Answer

  • [tex]a(t)=10\cos5t[/tex]
    [tex]v(t)=\displaystyle\int 10\cos5t\,\mathrm dt=2\sin5t+C_1[/tex]

    Since [tex]v(0)=-10[/tex], you have

    [tex]-10=2\sin0+C_1\implies C_1=-10[/tex]

    so

    [tex]v(t)=2\sin5t-10[/tex]

    Next,

    [tex]s(t)=\displaystyle\int(2\sin5t-10)\,\mathrm dt=-\dfrac25\cos5t-10t+C_2[/tex]

    Since [tex]s(0)=10[/tex], you get

    [tex]10=-\dfrac25\cos0-10\times0+C_2\implies C_2=\dfrac{52}5[/tex]

    and so

    [tex]s(t)=-\dfrac25\cos5t-10t+\dfrac{52}5[/tex]
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