Question

A particle has ~r(0) = (4 m) ˆ and ~v(0) = (2 m/s)ˆı. If its acceleration is constant and given by ~a = −(2 m/s 2 ) (ˆı +ˆ), at what time t does the particle first cross the x axis? Answer in units of s.

Answer 1

The particle has $\vec r(0)=(4\,\mathrm m)\,\vec\imath$ and $\vec v(0)=\left(2\frac{\rm m}{\rm s}\right)\,\vec\imath$, and is undergoing a constant acceleration of $\vec a=-\left(2\frac{\rm m}{\mathrm s^2}\right)(\vec\imath+\vec\jmath)$.

This means its position at time $t$ is given by the vector function,

$\vec r(t)=\vec r(0)+\vec v(0)t+\dfrac12\vec a t^2$

$\implies\vec r(t)=\left[4\,\mathrm m+\left(2\dfrac{\rm m}{\rm s}\right)t-\left(1\dfrac{\rm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath-\left(1\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath$

The particle crosses the x-axis when the $\vec\imath$ component is 0 for some time $t>0$, so we solve:

$4+2t-t^2=0\implies t^2-2t+1=5$

$\implies(t-1)^2=5$

$\implies t-1=\pm\sqrt5$

$\implies t=1\pm\sqrt5$

The negative square root introduces a negative solution that we throw out, leaving us with $\boxed{t=1+\sqrt5}$ or about 3.24 seconds after it starts moving.