a particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t…

a particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. find the velocity and speed (in m/s) when t = 4.\n\nf(t)=12 + \\frac{30}{t + 1}\n\nvelocity \n\nm/s\nspeed \n\nm/s

a particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. find the velocity and speed (in m/s) when t = 4.\n\nf(t)=12 + \\frac{30}{t + 1}\n\nvelocity \n\nm/s\nspeed \n\nm/s

Answer

Explanation:

Step1: Recall velocity formula

Velocity $v(t)$ is the derivative of position function $s(t)$. Given $s(t)=12+\frac{30}{t + 1}=12+30(t + 1)^{-1}$.

Step2: Differentiate using power - rule

The derivative of a constant is 0. For $y = 30(t + 1)^{-1}$, by the power - rule $\frac{d}{dt}(u^n)=nu^{n - 1}\frac{du}{dt}$, where $u=t + 1$ and $n=-1$. So $v(t)=s^\prime(t)=30\times(-1)\times(t + 1)^{-2}\times1=-\frac{30}{(t + 1)^2}$.

Step3: Find velocity at $t = 4$

Substitute $t = 4$ into $v(t)$. $v(4)=-\frac{30}{(4 + 1)^2}=-\frac{30}{25}=-\frac{6}{5}=-1.2$ m/s.

Step4: Find speed at $t = 4$

Speed is the absolute value of velocity. So speed at $t = 4$ is $|v(4)|=\left|-\frac{6}{5}\right|=\frac{6}{5}=1.2$ m/s.

Answer:

velocity: $-1.2$ m/s speed: $1.2$ m/s