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

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`.)

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) &= (d)/(d t)\left[x(t)\right] \\ &= (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) &= (d)/(d t)\left[v(t)\right] \\ &= (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`.