Given the position function, find
the speed.

Speed

Ask by Edwards Love. in the United States

Jan 29,2025

Question

Given the position function, find the speed. $$ \vec{r}(t)=t \vec{\imath}+(2 t-5) \vec{\jmath}+3 t \vec{k} $$ Speed $$ =\sqrt{[?]} $$

UpStudy AI · Accepted Answer

To find the **speed** of an object given its position vector function, follow these steps:

1. **Given Position Vector:**
   $$
   \vec{r}(t) = t\, \vec{\imath} + (2t - 5)\, \vec{\jmath} + 3t\, \vec{k}
   $$

2. **Find the Velocity Vector ($$\vec{v}(t)$$):**
   The velocity vector is the derivative of the position vector with respect to time $$t$$.
   $$
   \vec{v}(t) = \frac{d\vec{r}(t)}{dt} = \frac{d}{dt}\left( t\, \vec{\imath} + (2t - 5)\, \vec{\jmath} + 3t\, \vec{k} \right)
   $$
   $$
   \vec{v}(t) = 1\, \vec{\imath} + 2\, \vec{\jmath} + 3\, \vec{k}
   $$

3. **Calculate the Speed:**
   Speed is the magnitude of the velocity vector.
   $$
   \text{Speed} = \|\vec{v}(t)\| = \sqrt{(1)^2 + (2)^2 + (3)^2}
   $$
   $$
   \text{Speed} = \sqrt{1 + 4 + 9} = \sqrt{14}
   $$

**Final Answer:**
$$
\text{Speed} = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}
$$
Speed $$ = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14} $$

Given the position function, find
the speed.

Speed

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Given the position function, find the speed. Speed