Vector Problems ```Name: Unknown Status: N/A Age: N/A Location: N/A Country: N/A Date: Around 1993 ``` 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. Replies: 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 Click here to return to the Physics Archives

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