Question

the position of a particle moving along a coordinate line is s = √(8 + 4t), with s in meters and t in seconds. find the particles velocity and acceleration at t = 2 sec. the velocity at t = 2 sec is 1/2 m/sec. (simplify your answer. type an integer or a fraction.) the acceleration at t = 2 sec is m/sec². (simplify your answer. type an integer or a fraction.)

Explanation:

Step1: Find the velocity function

Velocity $v(t)$ is the derivative of the position - function $s(t)$. Given $s(t)=\sqrt{8 + 4t}=(8 + 4t)^{\frac{1}{2}}$. Using the chain - rule $\frac{d}{dt}(u^n)=nu^{n - 1}\cdot u'$, where $u = 8+4t$, $n=\frac{1}{2}$, and $u'=4$. So $v(t)=\frac{1}{2}(8 + 4t)^{-\frac{1}{2}}\cdot4=\frac{2}{\sqrt{8 + 4t}}$.

Step2: Calculate the velocity at $t = 2$

Substitute $t = 2$ into $v(t)$: $v(2)=\frac{2}{\sqrt{8+4\times2}}=\frac{2}{\sqrt{16}}=\frac{2}{4}=\frac{1}{2}$ m/sec.

Step3: Find the acceleration function

Acceleration $a(t)$ is the derivative of the velocity function $v(t)$. $v(t)=2(8 + 4t)^{-\frac{1}{2}}$. Using the chain - rule again, $a(t)=2\times(-\frac{1}{2})(8 + 4t)^{-\frac{3}{2}}\cdot4=-\frac{4}{(8 + 4t)^{\frac{3}{2}}}$.

Step4: Calculate the acceleration at $t = 2$

Substitute $t = 2$ into $a(t)$: $a(2)=-\frac{4}{(8 + 4\times2)^{\frac{3}{2}}}=-\frac{4}{16^{\frac{3}{2}}}=-\frac{4}{( \sqrt{16})^3}=-\frac{4}{64}=-\frac{1}{16}$ m/sec².

Answer:

The acceleration at $t = 2$ sec is $-\frac{1}{16}$ m/sec².