Question

question 1 - 1 point
the position in feet of an object moving in a straight line after t seconds is given by s(t)= - 4t^2 - 4. find its instantaneous velocity at t = 1 second.
provide your answer below:
v(1)=□

Explanation:

Step1: Recall velocity - position relation

The velocity function $v(t)$ is the derivative of the position function $s(t)$. Given $s(t)=-4t^{2}-4$, we use the power - rule for differentiation. The power - rule states that if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$.

Step2: Differentiate $s(t)$

For $s(t)=-4t^{2}-4$, the derivative of $-4t^{2}$ is $-4\times2t=-8t$ (using the power - rule with $a=-4$ and $n = 2$), and the derivative of the constant $-4$ is $0$. So, $v(t)=s^\prime(t)=-8t$.

Step3: Evaluate $v(t)$ at $t = 1$

Substitute $t = 1$ into $v(t)$. We get $v(1)=-8\times1=-8$.

Answer:

$-8$