Question

A particle moving along the x-axis has its velocity described by the function $v_x=2t^2$m/s, where t is in s. Its initial position is x0 = 1 m at t0 = 0 s. At t = 1 s what are the particle’s (a) position, (b) velocity, and (c) acceleration?

Short answer

(a) The position of the particle at $t=1s$is $1.67m$.

(b) The velocity of the particle at $t=1s$is $2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

(c) The acceleration of the particle at$t=1s$is$4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.

Step-by-step solution

Given information

The velocity of a particle is given by a function $v_x=2t^2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

The initial position of a particle is role="math" localid="1648217887311" $x_i=1m$.

Time$t_i=0s,t_f=1s$.

Part (a)The position of the particle at t=1s.

Find ${\int }_{t_i}^{t_f}{v}_{x}dt$.

Substitute known values.

$$\begin{gathered} {\int }_0^{1}2t^{2}dt=2{\left[\frac{t^3}{3}\right]}_0^1 \\ =2\left(\frac{1}{3}\right) \\ =\frac{2}{3} \end{gathered}$$

The position of the particle is given by$x_f=x_i+{\int }_{t_i}^{t_f}{v}_{x}dt$.

Substitute known values.

$$\begin{gathered} x_f=1+\frac{2}{3} \\ =1.67m \end{gathered}$$

Therefore, the position of the particle at$t=1s$is$1.67m$.

Part (b)The velocity of the particle at t=1 s.

To find the velocity of the particle at $t=1s$, substitute 1 for $t$into the given function.

role="math" localid="1648219607504" $$\begin{gathered} v\left(att=1s\right)=2{\left(1\right)}^2 \\ =2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Therefore, the velocity of the particle at$t=1s$is$2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Part (c)The acceleration of the particle at t=1s.

Acceleration is given by $a_x=\frac{d{v}_x}{dt}$.

Differentiate the given function with respect to $t$.

$$\begin{gathered} a_x=\frac{d\left(2t^2\right)}{dt} \\ =2\left(2t\right) \\ =4t \end{gathered}$$

Substitute 1 for $t$into the obtained expression.

$$\begin{gathered} a_x\left(att=1s\right)=4\left(1\right) \\ =4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \end{gathered}$$

Therefore, the acceleration of the particle is $4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.