An object moves from left to right along a parabolic shaped path y=(1/4)x^2. At the point (-2,1) the speed is 1. What normal acceleration is the object experiencing at that point?
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To determine the normal acceleration of an object at a given point along the path, we first need to find the equation for the velocity vector and then differentiate it to find the acceleration vector.
Step 1: Finding the equation for the velocity vector
The given parabolic path equation is y = (1/4)x^2. To find the velocity vector, we need to differentiate this equation with respect to time. However, since x and y represent position along the path, we need to differentiate them with respect to each other using the chain rule.
Differentiating y = (1/4)x^2 with respect to time (denoted by t), we have:
dy/dt = (1/4)(d/dt)(x^2)
Now, we need to find dx/dt, which represents the rate of change of x with respect to t. To do this, we can use the equation x = vt, where v represents the velocity along the x-axis.
Taking the derivative of the equation x = vt, we get:
dx/dt = v
Substituting this expression for dx/dt back into the previous equation, we have:
dy/dt = (1/4)(d/dt)(x^2)
dy/dt = (1/4)(2x)(dx/dt)
dy/dt = (1/2)x(dx/dt)
Hence, the equation for the velocity vector is:
v = (dx/dt)i + (1/2)x(dx/dt)j
Step 2: Evaluating the velocity vector
We're given that the speed is 1 at the point (-2,1). The speed is the magnitude of the velocity vector, so we can equate it to 1.
1 = sqrt((dx/dt)^2 + [(1/2)x(dx/dt)]^2)
Squaring both sides and rearranging, we get:
1 = (dx/dt)^2 + (1/4)x^2(dx/dt)^2
1 = (1 + (1/4)x^2)(dx/dt)^2
At the point (-2,1), we have:
1 = (1 + (1/4)(-2)^2)(dx/dt)^2
1 = (1 + 1/4)(dx/dt)^2
1 = (5/4)(dx/dt)^2
Solving for (dx/dt)^2, we find:
(dx/dt)^2 = 4/5
Taking the square root of both sides, we have:
dx/dt = ±sqrt(4/5)
dx/dt = ±(2/√5)
dx/dt = ±(2√5)/5
Step 3: Finding the acceleration vector
To find the acceleration vector, we need to take the derivative of the velocity vector with respect to time.
Differentiating v = (dx/dt)i + (1/2)x(dx/dt)j with respect to time (denoted by t), we have:
a = (d^2x/dt^2)i + x(d^2x/dt^2)j
Differentiating dx/dt = ±(2√5)/5 with respect to time, we get:
d^2x/dt^2 = 0 (since dx/dt is a constant at the given point)
Hence, the acceleration vector at the point (-2,1) is:
a = 0i + x(0)j
a = 0i
Therefore, the object is experiencing zero normal acceleration at the point (-2,1) along the parabolic path.