suppose the position of an object moving in a straight line is given by s(t) = t² + 3t + 4. find the…

Explanation:

Step1: Recall velocity - position relationship

The velocity function $v(t)$ is the derivative of the position function $s(t)$. Given $s(t)=t^{2}+3t + 4$, we use the power - rule for differentiation $\frac{d}{dt}(t^{n})=nt^{n - 1}$.

Step2: Differentiate $s(t)$

$v(t)=s^\prime(t)=\frac{d}{dt}(t^{2}+3t + 4)=\frac{d}{dt}(t^{2})+\frac{d}{dt}(3t)+\frac{d}{dt}(4)$. Using the power - rule, $\frac{d}{dt}(t^{2}) = 2t$, $\frac{d}{dt}(3t)=3$, and $\frac{d}{dt}(4)=0$. So $v(t)=2t + 3$.

Step3: Evaluate $v(t)$ at $t = 4$

Substitute $t = 4$ into $v(t)$. $v(4)=2\times4+3$. $v(4)=8 + 3=11$.

Answer:

$11$