Question

A particle’s acceleration is described by the function ax = 110 - t2 m/s2 , where t is in s. Its initial conditions are x0 = 0 m and v0x = 0 m/s at t = 0 s. a. At what time is the velocity again zero? b. What is the particle’s position at that time?

Short answer

Part (a)

The particle will have zero velocity again at $t=20s$.

Part (b)

The position of the particle at$t=20s$is$667m$.

Step-by-step solution

Given information

An acceleration of the particle is given by the function $a_x=\left(10-t\right)\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.

Initial conditions are $x_0=0m$,$v_{0x}=0m/s$,and $t=0s$.

Part (a)

The velocity is given by

$$\begin{gathered} v_f=v_{0x}+{\int }_{t_0}^{t_f}{a}_{x}dt \\ =v_{0x}+{\int }_{t_0}^{t_f}\left(10-t\right)dt \\ =v_{0x}+{\left[10t-\frac{t^2}{2}\right]}_{t_0}^{t_f} \end{gathered}$$

Substitute the known values

$$\begin{gathered} v_1=v_{0x}+{\left[10t-\frac{t^2}{2}\right]}_0^{t_1} \\ 0=0+\left(10t_1-\frac{t_1^2}{2}\right) \\ {t}_1^2-20t_1=0 \\ {t}_1=0sor{t}_1=20s \end{gathered}$$

Therefore, the particle will have zero velocity at $t_1=20s$.

Part (b)

The position of the particle is

$$\begin{gathered} x_f=x_0+{\int }_{t_0}^{t_f}{v}_{x}dt \\ =x_0+{\int }_{t_0}^{t_f}\left(10t-\frac{1}{2}{t}^2\right)dt \\ =x_0+{\left[5t^2-\frac{t^3}{6}\right]}_{t_0}^{t_f} \end{gathered}$$

Substitute the known values

$$\begin{gathered} x_1=x_0+{\left[5t^2-\frac{t^3}{6}\right]}_{t_0}^{t_1} \\ {x}_1=x_0+{\left[5t^2-\frac{t^3}{6}\right]}_0^{20} \\ =0+5{\left(20\right)}^2-\frac{{20}^3}{6} \\ =667m \end{gathered}$$

Therefore, the position of the particle at$t=20s$is$667m$.