Question

the position x of a watermelon as a function of time t is given by x(t)=-4t² - 6t - 8 where x is in m and t is in s measured by a stopwatch. part 1 what is the velocity v of the watermelon as a function of time? hint: assume that the units are m/s. there is no need to include them in your equation. use the following table as a reference. for use t t v=-8t - 6 part 2 at what time t₀ is the watermelon at rest? (negative values of t mean the watermelon is was at rest by that amount of time before the stopwatch was started.) t₀=number (rtol=0.05, atol=1e - 08) s

Explanation:

Step1: Recall velocity - position relation

Velocity $v(t)$ is the derivative of position $x(t)$ with respect to time $t$. Given $x(t)=-4t^{2}-6t - 8$, we use the power - rule for differentiation $\frac{d}{dt}(at^{n})=nat^{n - 1}$.

Step2: Differentiate each term

For the first term $-4t^{2}$, $\frac{d}{dt}(-4t^{2})=-4\times2t=-8t$. For the second term $-6t$, $\frac{d}{dt}(-6t)=-6$. For the third term $-8$ (a constant), $\frac{d}{dt}(-8) = 0$. So $v(t)=\frac{d}{dt}(-4t^{2}-6t - 8)=-8t-6$.

Step3: Find when the watermelon is at rest

The watermelon is at rest when $v(t)=0$. So we set $-8t-6 = 0$.

Step4: Solve for $t$

Add $6$ to both sides of the equation: $-8t=6$. Then divide both sides by $-8$ to get $t=-\frac{6}{8}=-\frac{3}{4}=- 0.75$.

Answer:

Part 1: $v(t)=-8t - 6$
Part 2: $t_0=-0.75$