Question

The position of a particle as a function of time is given by $x\left(t\right)=\ln (2t+3)$ for all positive $t$, where $x$ is in meters and $t$ is in seconds.
What is the particle's acceleration at $t=10sec$?

• A. $-\frac{1}{23}m/s^2$
• B. $-\frac{2}{23}m/s^2$
• C. $-\frac{4}{529}m/s^2$
• D. $-\frac{4}{23}m/s^2$
• E. $-\frac{1}{5}m/s^2$

Answer 1

The correct option is C $-\frac{4}{529}m/s^2$

Position of the particle $x\left(t\right)=\ln (2t+3)$ m

Differentiating w.r.t time we get its velocity $v\left(t\right)=\frac{d\left[x\right(t\left)\right]}{dt}=\frac{2}{2t+3}$ $m/s$

Differentiating again w.r.t time we get its acceleration $a\left(t\right)=\frac{d\left[v\right(t\left)\right]}{dt}=2\times \frac{-2}{(2t+3{)}^2}=\frac{-4}{(2t+3{)}^2}$ $m/s^2$

$⟹$ $a{|}_{t=10s}=\frac{-4}{(2\times 10+3{)}^2}=\frac{-4}{529}$ $m/s^2$