Position Physics Example 4

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Example 4

hard

A particle's position is

x(t) = t^3 - 6t^2 + 9t

(meters). At what times is the particle at position

x = 0

? What is the velocity at each of those times?

Solution

1
Set $x(t) = 0$ : $t^3 - 6t^2 + 9t = 0 \implies t(t^2 - 6t + 9) = 0 \implies t(t - 3)^2 = 0$ . So $t = 0 \text{ s}$ and $t = 3 \text{ s}$ .
2
Velocity: $v(t) = \frac{dx}{dt} = 3t^2 - 12t + 9$ . At $t = 0$ : $v(0) = 9 \text{ m/s}$ . At $t = 3$ : $v(3) = 27 - 36 + 9 = 0 \text{ m/s}$ .
3
The particle passes through the origin at $t = 0$ moving forward, and returns to the origin at $t = 3$ momentarily at rest.

Answer

t = 0 \text{ s } (v = 9 \text{ m/s}), \quad t = 3 \text{ s } (v = 0 \text{ m/s})

Setting the position function equal to a specific value finds the times when the particle is at that location. The velocity at each time tells us whether the particle is passing through or turning around.

About Position

The location of an object relative to a chosen reference point (origin), described using coordinates in a given reference frame.

Learn more about Position →

More Position Examples

Example 1 easy

An object starts at position [formula] and moves to [formula] along a straight line. What is the obj

Example 2 medium

A particle's position is given by [formula] (in meters, with [formula] in seconds). What is the part

Example 3 medium

An ant walks from position [formula] to [formula], then back to [formula]. What is the total distanc

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Related Concepts

Displacement Velocity