Vector Problems

Question: To any scientist it may concern:  I am in desperate need of 
physics help.  I am having serious problems with my Physics 103:  Mechanics 
class.  I have a problem with vectors in particular.  For example:  A particle 
leaves an origin with an initial velocity of v=3.0i (i being the x- component 
of velocity vector v).  It experiences a constant acceleration a=(-1.0i-o.5j), 
in m/s^2, (a) What is the velocity of the particle when it reaches its maximum 
x coordinate?   (b) Where is the particle at this time?  A detailed 
explanation of how to solve this problem and reasons for respective solution 
steps would be greatly appreciated.
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Okay, the important thing is to write down the equation of motion 
for each component, the x and the y, separately.  The definition of 
acceleration in the x direction is a_x = d2x / dt2 , or the second derivative 
of x with respect to time.  Since the acceleration is constant, if we 
integrate this equation, we get  dx/dt = a_x * t + (v_x)_0, where dx/dt is the 
velocity of the x coordinate and (v_x)_0 is the initial velocity in the x 
direction.  If we integrate again, we get x(t) = 0.5* a_x * t^2 + (v_x)_0 * t 
+ x_0, where x(t) is the x coordinate as a function of time and x_0 is the 
initial position.  Similarly,  y(t) = 0.5* a_y * t^2 + (v_y)_0 * t + y_0.  Now 
we plug in the initial conditions.  In the problem, we start at the origin, so 
x_0 = y_0 = 0.  The acceleration is (-1.0i,-0.5j) [the particle will be pulled 
down and to the left] and the initial velocity is (3.0i, 0.0j).  Substituting 
these numbers into the two equations, we get x(t) = -0.5 * t^2 + 3.0 * t [ a_x 
= -1.0,(v_x)_0 = -0.5, x_0 = 0] y(t) = -0.25* t^2    [ a_y = -0.5,(v_y)_0 = 
0.0, y_0 = 0] Now we can answer the questions.  (a) What is the velocity of 
the particle when it reaches its maximum x coordinate?  Well, it should be 
apparent that the velocity in the x coordinate will be zero when the particle 
reaches its maximum value of x.  But what is the y component of the velocity 
at this time?  Well, first we need to know at what TIME the particle is at its 
maximum x value.  To find this time, we differentiate x(t) and set the result 
equal to zero; dx/dt = - t + 3.0 = 0 ; therefore, t=3.0 is the time at which 
x(t) takes on it maximum value (this can be checked by plugging it into 
d2x/dt2 and checking that the answer is negative, i.e., x(t) is "concave down" 
at this critical point).  Next, we differentiate y(t), which results in dy/dt 
= -0.5* t.  Plugging in t=3, we get dy/dt = -1.5, and so the velocity of the 
particle when it reaches its maximum x coordinate is ...  (0.0i, -1.5j) m/sec.  
b) Where is the particle at this time?  Just plug this value of t into the two 
equation above, and get x(3) = -0.5 * 9 + 3.0 * 3 =  4.5 y(3) = -0.25 * 9  = -
2.25 and the position is (4.5i, -2.25j) meters.

Robert Topper
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