Questions: r = [4.0 cm + (25 cm/s^2) t^2] i + (5.0 cm / s) t j t=0 r=4.0 cm t=2.0

r = [4.0 cm + (25 cm/s^2) t^2] i + (5.0 cm / s) t j
t=0  r=4.0 cm
t=2.0
Transcript text: $\begin{array}{l}\vec{r}=\left[4.0 \mathrm{~cm}+\left(\frac{25 \mathrm{~cm}}{\mathrm{~s}^{2}}\right) t^{2}\right] i+(5.0 \mathrm{~cm} / \mathrm{s}) \mathrm{t} \hat{j} \\ t=0 \quad \vec{r}=4.0 \mathrm{~cm} \\ t=2.0\end{array}$
failed

Solution

failed
failed
Paso 1: Identificar la expresión de la posición en función del tiempo

La posición r\vec{r} está dada por: r=[4.0cm+(25cms2)t2]i^+(5.0cm/s)tj^ \vec{r} = \left[4.0 \, \text{cm} + \left(\frac{25 \, \text{cm}}{\text{s}^2}\right) t^2 \right] \hat{i} + (5.0 \, \text{cm/s}) t \hat{j}

Paso 2: Evaluar la posición en t=0t = 0

Para t=0t = 0: r(0)=[4.0cm+(25cms2)(0)2]i^+(5.0cm/s)(0)j^=4.0cmi^ \vec{r}(0) = \left[4.0 \, \text{cm} + \left(\frac{25 \, \text{cm}}{\text{s}^2}\right) (0)^2 \right] \hat{i} + (5.0 \, \text{cm/s}) (0) \hat{j} = 4.0 \, \text{cm} \hat{i}

Paso 3: Evaluar la posición en t=2.0t = 2.0 s

Para t=2.0t = 2.0 s: r(2.0)=[4.0cm+(25cms2)(2.0)2]i^+(5.0cm/s)(2.0)j^ \vec{r}(2.0) = \left[4.0 \, \text{cm} + \left(\frac{25 \, \text{cm}}{\text{s}^2}\right) (2.0)^2 \right] \hat{i} + (5.0 \, \text{cm/s}) (2.0) \hat{j} r(2.0)=[4.0cm+100cm]i^+10.0cmj^ \vec{r}(2.0) = \left[4.0 \, \text{cm} + 100 \, \text{cm} \right] \hat{i} + 10.0 \, \text{cm} \hat{j} r(2.0)=104.0cmi^+10.0cmj^ \vec{r}(2.0) = 104.0 \, \text{cm} \hat{i} + 10.0 \, \text{cm} \hat{j}

Respuesta Final

r(2.0)=104.0cmi^+10.0cmj^\boxed{\vec{r}(2.0) = 104.0 \, \text{cm} \hat{i} + 10.0 \, \text{cm} \hat{j}}

Was this solution helpful?
failed
Unhelpful
failed
Helpful