Question

A particle’s position along the x-axis is described byx(t) = A t + B t2,where t is in seconds, x is in meters, and the constants A and B are given below.Randomized VariablesA = -3.15 m/sB = 2.4 m/s21 what is the position of the particle when the velocity is zero.

Answer 1

Given data:

Position of the particle along the x-axis;

$x(t)=At+Bt^2$

Here, A and B are constants with A=-3.15 m/s and B=2.4 m/s².

Therefore,

$x(t)=-3.15t+2.4t^2$

The velocity of the particle is given as,

$v(t)=\frac{dx(t)}{dt}$

Substituting x(t),

$\begin{gathered} v(t)=\frac{d}{dt}(-3.15t+2.4t^2) \\ =\frac{d}{dt}(-3.15t)+\frac{d}{dt}(2.4t^2) \\ =-3.15+2\times2.4t \\ =-3.15+4.8t \end{gathered}$

The time when the velocity will be zero is calculated by substituting v(t)=0 in the above expresion,

$\begin{gathered} 0=-3.15+4.8t \\ 4.8t=3.15 \\ t=\frac{3.15}{4.8} \\ \approx0.66\text{ s} \end{gathered}$

Therefore, the position of the particle when the velocity is zero is calculated by substituting t=0.66 s in the equation for the position of the particle,

$\begin{gathered} x(0.66\text{ s})=-3.15\times0.66+2.4\times(0.66)^2 \\ \approx-1.03\text{ m} \end{gathered}$

Therefore, the position of the particle when the velocity is zero is -1.03 m (1.03 m to the left of its initial position).