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Name: Unknown
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Date: Around 1993

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.

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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