Calculus AP Question

**meg24209** · Nov 14th 2007, 02:59 PM

A particle moves along the x axis so that at time t its position is given by x(t)= t^3 - 6t^2 + 9t + 11
a) What is velocity of particle at t=0
b) During what time intervals is partical moving to the left?
c) What is the total distance traveled by partical from t=0 to t=2

**Soroban** · Nov 14th 2007, 03:46 PM

Hello, meg24209!

A particle moves along the x axis so that at time $t$ its position
is given by: . $x(t) \:= \:t^3 - 6t^2 + 9t + 11$

a) What is velocity of particle at $t=0$ ?

You're expected to know that the velocity is the first derivaitve.

$v(t) \;=\;x'(t) \;=\;3t^2-12t + 9$

At $t=0\!:\;\;v(0) \;=\;3\!\cdot\!0^2 - 12\!\cdot\!0 + 9 \;=\;9$

b) During what time intervals is particle moving to the left?

When is $v(t)$ negative?

We have: . $3t^2 - 12t + 9 \:<\:0\quad\Rightarrow\quad (t-1)(t-3) \:<\:0$

And we find that the velocity is negative for: . $1 \:<\:t\:<\:3$

c) What is the total distance traveled by particle from $t=0$ to $t=2$ ?

This could be considered a trick question . . .

At $t=0\!:\;x(0)\;=\;0^3-6\!\cdot\!0^2 + 9\!\cdot\!0 + 11 \;=\;11$

At $t=1\!:\;x(1)\;=\;1^3-6\!\cdot\!1^2 + 9\!\cdot\!1 + 11 \;=\;15$ . . . The particle has moved 4 units to the right.

At $t=2\!:\;x(2)\;=\;2^3-6\!\cdot2^2 + 9\!\cdot\!2 + 11\;=\;3$ . . . The particle has moved 12 units to the left.

Therefore, the particle travelled $4 + 12 \:=\:16$ units.

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