A particle travels along the xx-axis such that its velocity is given by v, left bracket, t, right bracket, equals, t, start superscript, 1, point, 5, end superscript, sine, left bracket, 2, t, plus, 5, right bracket, .v(t)=t 1.5 sin(2t+5). What is the acceleration of the particle at time t, equals, 5, question markt=5? You may use a calculator and round your answer to the nearest thousandth.
Question
A particle travels along the xx-axis such that its velocity is given by v, left bracket, t, right bracket, equals, t, start superscript, 1, point, 5, end superscript, sine, left bracket, 2, t, plus, 5, right bracket, .v(t)=t 1.5 sin(2t+5). What is the acceleration of the particle at time t, equals, 5, question markt=5? You may use a calculator and round your answer to the nearest thousandth.
Solution
The acceleration of a particle at a given time is the derivative of its velocity at that time.
Given the velocity function v(t) = t^1.5 * sin(2t + 5), we need to find its derivative to get the acceleration function a(t).
The derivative of this function can be found using the product rule and the chain rule.
The product rule states that the derivative of two functions multiplied together is the derivative of the first function times the second function plus the first function times the derivative of the second function.
The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.
Applying these rules, we get:
a(t) = dv/dt = 1.5t^0.5 * sin(2t + 5) + t^1.5 * cos(2t + 5) * 2
Now, we substitute t = 5 into the acceleration function to find the acceleration at that time:
a(5) = 1.55^0.5 * sin(25 + 5) + 5^1.5 * cos(2*5 + 5) * 2
You can use a calculator to compute this expression and round your answer to the nearest thousandth.
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