We need to determine the velocity and acceleration vectors, then calculate the velocity's magnitude and direction at t = 2.0 s t = 2.0\,\text{s} t = 2.0 s
Concept: Derivative
The instantaneous velocity v ⃗ \vec{v} v r ⃗ \vec{r} r a ⃗ \vec{a} a
v ⃗ ( t ) = d r ⃗ d t , a ⃗ ( t ) = d v ⃗ d t \vec{v}(t) = \frac{d\vec{r}}{dt},
\quad
\vec{a}(t) = \frac{d\vec{v}}{dt} v ( t ) = d t d r , a ( t ) = d t d v The magnitude of velocity is given by the Euclidean norm of its components, and the direction θ \theta θ x y xy x y
∣ v ⃗ ∣ = v x 2 + v y 2 + v z 2 , θ = arctan ( v y v x ) |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2},
\quad
\theta = \arctan\left(\frac{v_y}{v_x}\right) ∣ v ∣ = v x 2 + v y 2 + v z 2 , θ = arctan ( v x v y ) Given:
Position vector: r ⃗ ( t ) = 3.0 t i ^ − 2.0 t 2 j ^ + 4.0 k ^ m \vec{r}(t) = 3.0 t \hat{i} - 2.0 t^{2} \hat{j} + 4.0 \hat{k}\,\text{m} r ( t ) = 3.0 t i ^ − 2.0 t 2 j ^ + 4.0 k ^ m
Time of interest: t = 2.0 s t = 2.0\,\text{s} t = 2.0 s
Solving:
Find the velocity vector v ⃗ ( t ) \vec{v}(t) v ( t )
Differentiate each component of r ⃗ ( t ) \vec{r}(t) r ( t ) t t t
v ⃗ ( t ) = d d t ( 3.0 t ) i ^ + d d t ( − 2.0 t 2 ) j ^ + d d t ( 4.0 ) k ^ = 3.0 i ^ − 4.0 t j ^ + 0 k ^ = 3.0 i ^ − 4.0 t j ^ m ⋅ s − 1 \begin{aligned}
\vec{v}(t) &= \frac{d}{dt}(3.0 t) \hat{i} + \frac{d}{dt}(-2.0 t^{2}) \hat{j} + \frac{d}{dt}(4.0) \hat{k} \\
&= 3.0 \hat{i} - 4.0 t \hat{j} + 0 \hat{k} \\
&= 3.0 \hat{i} - 4.0 t \hat{j}\,\text{m}\cdot\text{s}^{-1}
\end{aligned} v ( t ) = d t d ( 3.0 t ) i ^ + d t d ( − 2.0 t 2 ) j ^ + d t d ( 4.0 ) k ^ = 3.0 i ^ − 4.0 t j ^ + 0 k ^ = 3.0 i ^ − 4.0 t j ^ m ⋅ s − 1
Find the acceleration vector a ⃗ ( t ) \vec{a}(t) a ( t )
Differentiate the velocity vector v ⃗ ( t ) \vec{v}(t) v ( t ) t t t
a ⃗ ( t ) = d d t ( 3.0 ) i ^ + d d t ( − 4.0 t ) j ^ = 0 i ^ − 4.0 j ^ = − 4.0 j ^ m ⋅ s − 2 \begin{aligned}
\vec{a}(t) &= \frac{d}{dt}(3.0) \hat{i} + \frac{d}{dt}(-4.0 t) \hat{j} \\
&= 0 \hat{i} - 4.0 \hat{j} \\
&= -4.0 \hat{j}\,\text{m}\cdot\text{s}^{-2}
\end{aligned} a ( t ) = d t d ( 3.0 ) i ^ + d t d ( − 4.0 t ) j ^ = 0 i ^ − 4.0 j ^ = − 4.0 j ^ m ⋅ s − 2
Calculate velocity at t = 2.0 s t = 2.0\,\text{s} t = 2.0 s
Substitute t = 2.0 t = 2.0 t = 2.0 v ⃗ ( t ) \vec{v}(t) v ( t )
v ⃗ ( 2.0 ) = 3.0 i ^ − 4.0 ( 2.0 ) j ^ = 3.0 i ^ − 8.0 j ^ m ⋅ s − 1 \begin{aligned}
\vec{v}(2.0) &= 3.0 \hat{i} - 4.0(2.0) \hat{j} \\
&= 3.0 \hat{i} - 8.0 \hat{j}\,\text{m}\cdot\text{s}^{-1}
\end{aligned} v ( 2.0 ) = 3.0 i ^ − 4.0 ( 2.0 ) j ^ = 3.0 i ^ − 8.0 j ^ m ⋅ s − 1
Calculate the magnitude of velocity at t = 2.0 s t = 2.0\,\text{s} t = 2.0 s
∣ v ⃗ ∣ = ( 3.0 ) 2 + ( − 8.0 ) 2 = 9.0 + 64.0 = 73.0 ≈ 8.54 m ⋅ s − 1 \begin{aligned}
|\vec{v}| &= \sqrt{(3.0)^2 + (-8.0)^2} \\
&= \sqrt{9.0 + 64.0} \\
&= \sqrt{73.0} \\
&\approx 8.54\,\text{m}\cdot\text{s}^{-1}
\end{aligned} ∣ v ∣ = ( 3.0 ) 2 + ( − 8.0 ) 2 = 9.0 + 64.0 = 73.0 ≈ 8.54 m ⋅ s − 1
Calculate the direction of velocity at t = 2.0 s t = 2.0\,\text{s} t = 2.0 s
The angle θ \theta θ x x x
θ = arctan ( − 8.0 3.0 ) ≈ − 69.4 ∘ \begin{aligned}
\theta &= \arctan\left(\frac{-8.0}{3.0}\right) \\
&\approx -69.4^{\circ}
\end{aligned} θ = arctan ( 3.0 − 8.0 ) ≈ − 69. 4 ∘ This angle is 69.4 ∘ 69.4^{\circ} 69. 4 ∘ x x x
Answer
(a)
v ⃗ ( t ) = 3.0 i ^ − 4.0 t j ^ m ⋅ s − 1 \vec{v}(t) = 3.0 \hat{i} - 4.0 t \hat{j}\,\text{m}\cdot\text{s}^{-1} v ( t ) = 3.0 i ^ − 4.0 t j ^ m ⋅ s − 1 a ⃗ ( t ) = − 4.0 j ^ m ⋅ s − 2 \vec{a}(t) = -4.0 \hat{j}\,\text{m}\cdot\text{s}^{-2} a ( t ) = − 4.0 j ^ m ⋅ s − 2 (b) Magnitude:
8.54 m ⋅ s − 1 8.54\,\text{m}\cdot\text{s}^{-1} 8.54 m ⋅ s − 1 Direction:
− 69.4 ∘ with the x -axis -69.4^{\circ} \text{ with the } x\text{-axis} − 69. 4 ∘ with the x -axis