Question

A particle’s velocity is described by the function vx = kt 2 m/s, where k is a constant and t is in s. The particle’s position at t0 = 0 s is x0 = - 9.0 m. At t1 = 3.0 s, the particle is at x1 = 9.0 m. Determine the value of the constant k. Be sure to include the proper units

Short answer

The value of constant$k$is 2.

Step-by-step solution

Given information

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

The position of the particle at $t_0=0s$is $x_0=-9.0m$.

The position of the particle at$t_1=3.0s$is$x_1=9.0m$.

The value of constant k.

The position of a particle is given by

$x_f=x_0+{\int }_{t_0}^{t_f}{v}_{x}dt$

here,

role="math" localid="1648490216091" $$\begin{gathered} x_1=x_0+{\int }_{t_0}^{t_1}{v}_{x}dt \\ =x_0+{\int }_{t_0}^{t_1}k{t}^{2}dt \\ =x_0+{\left[\frac{1}{3}k{t}^3\right]}_{t_0}^{t_1} \\ =x_0+\frac{1}{3}k{t}_1^3-\frac{1}{3}k{t}_0^3 \end{gathered}$$

Substitute known values

$$\begin{gathered} 9=-9+\frac{1}{3}k{\left(3\right)}^3-\frac{1}{3}k{\left(0\right)}^3 \\ 9=-9+9k \\ 9k=18 \\ k=2 \end{gathered}$$

Therefore, the value of constant$k$is 2.