A particle moves a distance x in time t according to equation x = (t + 5)–1. The acceleration of the particle is proportional to :(velocity)3/2 (distance)2 (distance)–2(velocity)2/3
Question
A particle moves a distance x in time t according to equation x = (t + 5)–1. The acceleration of the particle is proportional to :(velocity)3/2 (distance)2 (distance)–2(velocity)2/3
Solution
The given equation is x = (t + 5)–1.
First, we need to find the velocity (v) which is the first derivative of the displacement (x) with respect to time (t).
Then, we need to find the acceleration (a) which is the first derivative of the velocity (v) with respect to time (t).
After finding the velocity and acceleration, we can then determine which of the given options the acceleration is proportional to.
Let's start:
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Differentiate x = (t + 5)–1 with respect to t to find the velocity:
dx/dt = v = -1/(t + 5)^2
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Differentiate v = -1/(t + 5)^2 with respect to t to find the acceleration:
dv/dt = a = 2/(t + 5)^3
From the above, we can see that the acceleration is not proportional to any of the given options. The acceleration is proportional to the inverse cube of (t + 5), not the square or cube root of the velocity or distance.
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