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 $delta t
ightarrow0$. 1. $x(t)=x_0 + v_0t+\frac{at^2}{2}$ 2. $x(t)=\frac{at^2}{2}+\frac{bt^3}{3}$ 3. $x(t)=v_0t+\frac{pt^6}{6}$ 4. $x(t)=\frac{qt^5}{5}+\frac{rt^4}{4}$

Explanation:

Step1: Recall the definition of instantaneous velocity

Instantaneous velocity $v(t)=\lim_{\Delta t
ightarrow0}\frac{x(t + \Delta t)-x(t)}{\Delta t}$, which is equivalent to the derivative of the position - function $x(t)$ with respect to time $t$, i.e., $v(t)=x^\prime(t)$.

Step2: Differentiate $x(t)=x_0 + v_0t+\frac{1}{2}at^{2}$

Using the power - rule of differentiation $\frac{d}{dt}(t^n)=nt^{n - 1}$, we have:
$x^\prime(t)=\frac{d}{dt}(x_0)+\frac{d}{dt}(v_0t)+\frac{d}{dt}(\frac{1}{2}at^{2})$. Since $\frac{d}{dt}(x_0) = 0$ (derivative of a constant), $\frac{d}{dt}(v_0t)=v_0$ and $\frac{d}{dt}(\frac{1}{2}at^{2})=at$, then $v(t)=v_0+at$.

Step3: Differentiate $x(t)=\frac{1}{2}at^{2}+\frac{1}{3}bt^{3}$

Using the power - rule $\frac{d}{dt}(t^n)=nt^{n - 1}$, we get $x^\prime(t)=\frac{d}{dt}(\frac{1}{2}at^{2})+\frac{d}{dt}(\frac{1}{3}bt^{3})$. So $x^\prime(t)=at + bt^{2}$.

Step4: Differentiate $x(t)=v_0t+\frac{1}{6}pt^{3}$

Using the power - rule, $\frac{d}{dt}(v_0t)=v_0$ and $\frac{d}{dt}(\frac{1}{6}pt^{3})=\frac{1}{2}pt^{2}$, then $x^\prime(t)=v_0+\frac{1}{2}pt^{2}$.

Step5: Differentiate $x(t)=\frac{1}{4}st^{4}+\frac{1}{5}ct^{5}$

Using the power - rule, $\frac{d}{dt}(\frac{1}{4}st^{4})=st^{3}$ and $\frac{d}{dt}(\frac{1}{5}ct^{5})=ct^{4}$, then $x^\prime(t)=st^{3}+ct^{4}$.

Answer:

1. $v(t)=v_0 + at$
2. $v(t)=at+bt^{2}$
3. $v(t)=v_0+\frac{1}{2}pt^{2}$
4. $v(t)=st^{3}+ct^{4}$