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 1/5 m/sec. (simplify your answer. type an integer or a fraction.) the acceleration at t = 6 sec is m/sec². (simplify your answer. type an integer or a fraction.)

Explanation:

Step1: Find the velocity function

The 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 $\frac{d}{dt}[u^n]=n\cdot u^{n - 1}\cdot u'$, where $u = 76+4t$ and $n=\frac{1}{2}$, $u' = 4$. So $v(t)=\frac{1}{2}(76 + 4t)^{-\frac{1}{2}}\cdot4=\frac{2}{\sqrt{76 + 4t}}$.

Step2: Calculate the velocity at $t = 6$

Substitute $t = 6$ into $v(t)$: $v(6)=\frac{2}{\sqrt{76+4\times6}}=\frac{2}{\sqrt{76 + 24}}=\frac{2}{\sqrt{100}}=\frac{2}{10}=\frac{1}{5}$ m/s.

Step3: Find the acceleration function

The 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 $u = 76+4t$, $n=-\frac{1}{2}$, $u' = 4$. Then $a(t)=2\times(-\frac{1}{2})(76 + 4t)^{-\frac{3}{2}}\cdot4=-\frac{4}{(76 + 4t)^{\frac{3}{2}}}$.

Step4: Calculate the acceleration at $t = 6$

Substitute $t = 6$ into $a(t)$: $a(6)=-\frac{4}{(76+4\times6)^{\frac{3}{2}}}=-\frac{4}{100^{\frac{3}{2}}}=-\frac{4}{1000}=-\frac{1}{250}$ m/s².

Answer:

The acceleration at $t = 6$ sec is $-\frac{1}{250}$ m/s².