Question

for the following positions $x(t)$: find the instantaneous velocity $v(t)$. remember that the instantaneous velocity is the slope of $x(t)$: $(x(t + \delta t)-x(t))/(\delta t)$ as we let $\delta t\to0$.

1. $x(t)=x_0 + v_0t+\frac{at^{2}}{2}$
2. $x(t)=\frac{at^{2}}{2}+\frac{jt^{3}}{3}$
3. $x(t)=v_0t+\frac{pt^{6}}{6}$
4. $x(t)=\frac{qt^{5}}{5}+\frac{st^{4}}{4}$

Explanation:

Step1: Recall the definition of instantaneous velocity

The instantaneous velocity $v(t)$ is given by $v(t)=\lim_{\Delta t
ightarrow0}\frac{x(t + \Delta t)-x(t)}{\Delta t}$, which is the derivative of the position - function $x(t)$ with respect to time $t$. We will use the power - rule of differentiation $\frac{d}{dt}(t^n)=nt^{n - 1}$.

For $x(t)=x_0 + v_0t+\frac{1}{2}at^2$:

Step1: Find $x(t+\Delta t)$

$x(t+\Delta t)=x_0 + v_0(t + \Delta t)+\frac{1}{2}a(t+\Delta t)^2=x_0 + v_0t+v_0\Delta t+\frac{1}{2}a(t^2 + 2t\Delta t+\Delta t^2)=x_0 + v_0t+v_0\Delta t+\frac{1}{2}at^2+at\Delta t+\frac{1}{2}a\Delta t^2$

Step2: Calculate $\frac{x(t+\Delta t)-x(t)}{\Delta t}$

$x(t+\Delta t)-x(t)=v_0\Delta t+at\Delta t+\frac{1}{2}a\Delta t^2$. Then $\frac{x(t+\Delta t)-x(t)}{\Delta t}=v_0 + at+\frac{1}{2}a\Delta t$.

Step3: Take the limit as $\Delta t

ightarrow0$
$v(t)=\lim_{\Delta t
ightarrow0}\frac{x(t+\Delta t)-x(t)}{\Delta t}=v_0 + at$

For $x(t)=\frac{1}{2}at^2+\frac{1}{3}jt^3$:

Step1: Find $x(t+\Delta t)$

$x(t+\Delta t)=\frac{1}{2}a(t + \Delta t)^2+\frac{1}{3}j(t+\Delta t)^3=\frac{1}{2}a(t^2 + 2t\Delta t+\Delta t^2)+\frac{1}{3}j(t^3+3t^2\Delta t + 3t\Delta t^2+\Delta t^3)=\frac{1}{2}at^2+at\Delta t+\frac{1}{2}a\Delta t^2+\frac{1}{3}jt^3+jt^2\Delta t+jt\Delta t^2+\frac{1}{3}j\Delta t^3$

Step2: Calculate $\frac{x(t+\Delta t)-x(t)}{\Delta t}$

$x(t+\Delta t)-x(t)=at\Delta t+\frac{1}{2}a\Delta t^2+jt^2\Delta t+jt\Delta t^2+\frac{1}{3}j\Delta t^3$. Then $\frac{x(t+\Delta t)-x(t)}{\Delta t}=at+\frac{1}{2}a\Delta t+jt^2+jt\Delta t+\frac{1}{3}j\Delta t^2$

Step3: Take the limit as $\Delta t

ightarrow0$
$v(t)=\lim_{\Delta t
ightarrow0}\frac{x(t+\Delta t)-x(t)}{\Delta t}=at + jt^2$

For $x(t)=\frac{1}{4}st^4+\frac{1}{5}ct^5$:

Step1: Find $x(t+\Delta t)$

$x(t+\Delta t)=\frac{1}{4}s(t + \Delta t)^4+\frac{1}{5}c(t+\Delta t)^5$. Using the binomial expansion $(A + B)^n=\sum_{k = 0}^{n}\binom{n}{k}A^{n - k}B^{k}$, we have $x(t+\Delta t)=\frac{1}{4}s(t^4+4t^3\Delta t + 6t^2\Delta t^2+4t\Delta t^3+\Delta t^4)+\frac{1}{5}c(t^5+5t^4\Delta t+10t^3\Delta t^2 + 10t^2\Delta t^3+5t\Delta t^4+\Delta t^5)$

Step2: Calculate $\frac{x(t+\Delta t)-x(t)}{\Delta t}$

After simplifying $x(t+\Delta t)-x(t)$ and dividing by $\Delta t$, we get $\frac{x(t+\Delta t)-x(t)}{\Delta t}=st^3 + \text{(terms with }\Delta t\text{)}$

Step3: Take the limit as $\Delta t

ightarrow0$
$v(t)=\lim_{\Delta t
ightarrow0}\frac{x(t+\Delta t)-x(t)}{\Delta t}=st^3+ct^4$

For $x(t)=v_0t+\frac{1}{6}pt^6$:

Step1: Find $x(t+\Delta t)$

$x(t+\Delta t)=v_0(t + \Delta t)+\frac{1}{6}p(t+\Delta t)^6$. Expanding $(t+\Delta t)^6$ using the binomial theorem $(a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}$, we have $x(t+\Delta t)=v_0t+v_0\Delta t+\frac{1}{6}p(t^6+6t^5\Delta t + 15t^4\Delta t^2+20t^3\Delta t^3+15t^2\Delta t^4+6t\Delta t^5+\Delta t^6)$

Step2: Calculate $\frac{x(t+\Delta t)-x(t)}{\Delta t}$

$x(t+\Delta t)-x(t)=v_0\Delta t+\frac{1}{6}p(6t^5\Delta t+\text{(higher - order terms in }\Delta t\text{)})$. Then $\frac{x(t+\Delta t)-x(t)}{\Delta t}=v_0+pt^5+\text{(terms with }\Delta t\text{)}$

Step3: Take the limit as $\Delta t

ightarrow0$
$v(t)=\lim_{\Delta t
ightarrow0}\frac{x(t+\Delta t)-x(t)}{\Delta t}=v_0+pt^5$

Answer:

1. $v(t)=v_0 + at$
2. $v(t)=at + jt^2$
3. $v(t)=st^3+ct^4$
4. $v(t)=v_0+pt^5$