Question

. A particle moves on a line away from its initial position so that after t seconds its distance is s = 3t^2+tmeters from its initial position. (a) At what time does the particle have a velocity of 25 m/s? (b) Is the acceleration ever 0? Why or why/not? Explain your answer
pls solve and explain fast​

Answer 1

Answer:

The velocity of this particle is $25\; {\rm m \cdot s^{-1}}$ at $t = 4\; {\rm s}$.

Acceleration is constantly $6\; {\rm m\cdot s^{-2}}$ (and thus is never $0$.)

Step-by-step explanation:

The distance between the particle and the initial position is the displacement of this particle. Let $x(t)$ denote the displacement (in meters) of this particle at time $t$ (in seconds.) The question states that $x(t) = 3\, t^{2} + t$.

Differentiate displacement $x(t)$ with respect to time $t$ to find the velocity $v(t)$ (in ${\rm m \cdot s^{-1}}$) of this particle:

$\begin{aligned}v(t) &= \frac{d}{d t}\left[x(t)\right] \\ &= \frac{d}{d t}\left[3\, t^{2} + t\right] \\ &= 6\, t + 1 \end{aligned}$.

Set velocity to $25\; {\rm m\cdot s^{-1}}$ and solve for time $t$ (in seconds):

$v(t) = 25$.

$6\, t + 1 = 25$.

$t = 4$.

Thus, the velocity of this particle is $25\; {\rm m \cdot s^{-1}}$ at $t = 4\; {\rm s}$.

Differentiate velocity $v(t)$ with respect to time $t$ to find the acceleration (in ${\rm m\cdot s^{-2}}$) of this particle:

$\begin{aligned}a(t) &= \frac{d}{d t}\left[v(t)\right] \\ &= \frac{d}{d t}\left[6\, t + 1\right] \\ &= 6\end{aligned}$.

In other words, the acceleration of this particle is constantly equal to $6\; {\rm m\cdot s^{-2}}$. Hence, the acceleration of this particle is never $0$.