Applications of Calculus Question (1 Viewer)

rand_althor · Aug 16, 2015

$$A moving particle has an acceleration $a$ metres per second per second at time $t$ seconds $(t\geq0)$. The acceleration of the particle is given by $a=2\pi^2\cos{(\pi t)}$. At time $t=0$, the object is at the point $x=0$ and has a velocity $v=\pi$ metres per second. \\(c) Show that the displacement, $x$, is given by $x=-2\cos{(\pi t)}+\pi t+3$ for $t\geq0$.\\(d) Show that the maximum displacement of $x$ is given by $x=(3+\sqrt{3})+\frac{19\pi}{6}$, for $0\leq t \leq4$.$

My answer for (c) has a +2 at the end rather than a +3. Can't find a mistake in my working out, so if someone could please confirm what the right answer is, I'd appreciate it.

rand_althor · Aug 16, 2015

Nevermind, subbing in t=0 to x doesn't give x=0, so I'll assume the question is wrong.

InteGrand · Aug 16, 2015

rand_althor said:
$$A moving particle has an acceleration $a$ metres per second per second at time $t$ seconds $(t\geq0)$. The acceleration of the particle is given by $a=2\pi^2\cos{(\pi t)}$. At time $t=0$, the object is at the point $x=0$ and has a velocity $v=\pi$ metres per second. \\(c) Show that the displacement, $x$, is given by $x=-2\cos{(\pi t)}+\pi t+3$ for $t\geq0$.\\(d) Show that the maximum displacement of $x$ is given by $x=(3+\sqrt{3})+\frac{19\pi}{6}$, for $0\leq t \leq4$.$

My answer for (c) has a +2 at the end rather than a +3. Can't find a mistake in my working out, so if someone could please confirm what the right answer is, I'd appreciate it.

If it's a +3, then the initial conditions of x = 0 at t = 0 aren't satisfied (it would be x = 1 instead). So maybe the Q has a typo.

Kaido · Aug 16, 2015

c) take the integral twice, making sure you solve for +c each time by subbing known variables.
d) let v=0 , solve for t, plug it in a to test for max. sub it back into x to get ans

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Applications of Calculus Question (1 Viewer)

rand_althor

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rand_althor

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InteGrand

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Kaido

be.

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