Question

A particle moves in a way such that its position can be expressed with $x\left(t\right)=t^2+t-3$ and with $y\left(t\right)=t^3-\frac{1}{t-1}$, with being time in seconds. $x\left(t\right)$ and $y\left(t\right)$ are both in meters.
Initially, how far away is the particle from the origin?

• A. $0.00m$
• B. $1.00m$
• C. $1.56m$
• D. $2.83m$
• E. $3.16m$

Answer 1

The correct option is E $3.16m$

X-coordinate of the particle, $x\left(t\right)=t^2+t-3$ m

X-coordinate of the particle at $t=0$ s, $x_o=0+0-3=-3$ m

Y-coordinate of the particle $y\left(t\right)=t^3-\frac{1}{t-1}$ m

Y-coordinate of the particle at $t=0$ s, $y_o=0-\frac{1}{0-1}=1$ m

Thus distance of particle from the origin $D=\sqrt{(-3{)}^2+1^2}=\sqrt{10}=3.16$ m