Question

the position of a particle moving along a coordinate line is s = √(76 + 4t), with s in meters and t in seconds. find the particles velocity and acceleration at t = 6 sec
the velocity at t = 6 sec is m/sec.
(simplify your answer. type an integer or a fraction.)

Explanation:

Step1: Recall velocity - derivative of position

Velocity $v(t)$ is the derivative of the position function $s(t)$. Given $s(t)=\sqrt{76 + 4t}=(76 + 4t)^{\frac{1}{2}}$. Using the chain - rule, if $y = u^{\frac{1}{2}}$ and $u=76 + 4t$, then $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dt}=4$. So $v(t)=\frac{ds}{dt}=\frac{1}{2}(76 + 4t)^{-\frac{1}{2}}\times4=\frac{2}{\sqrt{76 + 4t}}$.

Step2: Calculate velocity at $t = 6$

Substitute $t = 6$ into the velocity function. When $t = 6$, $76+4t=76 + 4\times6=76 + 24 = 100$. Then $v(6)=\frac{2}{\sqrt{100}}=\frac{2}{10}=\frac{1}{5}$.

Step3: Recall acceleration - derivative of velocity

Acceleration $a(t)$ is the derivative of the velocity function $v(t)$. Since $v(t)=2(76 + 4t)^{-\frac{1}{2}}$, using the chain - rule again. Let $y = 2u^{-\frac{1}{2}}$ and $u = 76+4t$, $\frac{dy}{du}=-u^{-\frac{3}{2}}$ and $\frac{du}{dt}=4$. So $a(t)=\frac{dv}{dt}=-2(76 + 4t)^{-\frac{3}{2}}\times4=-\frac{8}{(76 + 4t)^{\frac{3}{2}}}$.

Step4: Calculate acceleration at $t = 6$

Substitute $t = 6$ into the acceleration function. When $t = 6$, $76 + 4t=100$. Then $a(6)=-\frac{8}{100^{\frac{3}{2}}}=-\frac{8}{1000}=-\frac{1}{125}$.

Answer:

The velocity at $t = 6$ sec is $\frac{1}{5}$ m/sec.
The acceleration at $t = 6$ sec is $-\frac{1}{125}$ m/sec².