Answer to Question #95008 in Classical Mechanics for Thomas Clifford

Question

A particle is moving in x-y plane. It's initial velocity & acceleration are u=(4î+5ĵ) m/s and a=(2î-4ĵ) m/s^2. Find
(1) the time when the particle will cross the x axis
(2) x coordinate of the particle at that instant
(3) velocity of the particle at this instant
Initial coordinates of particle are (4m,10m)

Answer 1

Question (a)

The cutting of the X axis occurs when the Y coordinate is equal to zero.

The vertical position is given by:$Y=Y_{o}+V_{oy}*t+\frac{1}{2}a_{y}t^{2}$

Where:

• Initial position. $Y_{o}=10m$
• Initial velocity $V_{oy}=+5\frac{m}{s}$
• Vertical acceleration. a$a_{y}=-4\frac{m}{s^{2}}$

Evaluating at Y = 0.

$0=10+5t-\frac{1}{2}4t^{2}\\ -2t^{2}+5t+10=0$

A second grade equation.$-2t^{2}+5t+10=0->ax^{2}+bx+c=0$

Coefficients $a=-2;b=5;c=10$

Solving the second degree equation.

$t=\frac{-b\pm \sqrt{b^{2}-4*a*c}}{2*a} \\t=\frac{-(5)\pm \sqrt{(5)^{2}-4*-2*10}}{2*-2}\\ t=\frac{-5 \pm 10.2}{-4}$

Solution 1

$t=\frac{-5 + 10.2}{-4} \\t=-1.25s$

Solution 2

$t=\frac{-5 - 10}{-4} \\t=3.75s$

the time when the particle will cross the x axis 

$\boxed{t=3.75s}$

Question (b)

The position of the particle is given by:

$\vec{r(t)}=\vec{r}_{o}+\vec{V_{o}}*t+\frac{1}{2}\vec{a}t^{2}$

Where:

• initial velocity $\vec{V_{0}}=(4i+5j)\frac{m}{s}$
• The acceleration is $\vec{a}=(2i-4j)\frac{m}{s^{2}}$
• Initial position $\vec{r_{0}}=(4i+10j)m$
• time $t=3.75s$

Numerically evaluating

$\vec{r(t)}=(4i+10j)m+(4i+5j)\frac{m}{s}*3.75s+\frac{1}{2}*(2i-4j)\frac{m}{s^{2}}*(3.75s)^{2}$

$\vec{r_{o}}=(4i+10j)m+(15i+18j)m+(14i-28j)m$

$\vec{r_{o}}=(33i+0j)m$

The position at that moment is:

$\boxed{\vec{r_{o}}=(33i+0j)m \text{ or } \vec{r_{o}}=(33m,0m)}$

Question (c)

The velocity of the particle is given by:

$\vec{V(t)}=\vec{V_{o}}+\vec{a}*t$

Where:

• The initial velocity$\vec{V_{0}}=(4i+5j)\frac{m}{s}$
• Acceleration $\vec{a}=(2i-4j)\frac{m}{s^{2}}$
• Time t=3.75s

Numerically evaluating

$\vec{V(t)}=(4i+5j)\frac{m}{s} +(2i-4j)\frac{m}{s^{2}}*3.75s \\\vec{V(t)}=(4i+5j)\frac{m}{s}+(5.5i-15j)\frac{m}{s}\\\vec{V(t)}=(11.5i-10j)\frac{m}{s}$

The velocity at that moment is

$\boxed{vec{V(t)}=(11.5i-10j)\frac{m}{s} }$