Acceleration to position Given the following acceleration func...

Question

Acceleration to position Given the following acceleration functions of an object moving along a line, find the position function with the given initial velocity and position.

a(t) = 2 + 3 sin t; v(0) = 1, s(0) = 10

Accepted Answer

Welcome back, everyone. Given A of T equals 3 + 2 cosine T with V of 0 equals -1 and S 0 equals 5, find the position functions of T. So we are provided with the acceleration function. What we want to do is simply recall that the velocity function is the integral of the acceleration function A of TDT, and then if we want to identify the position function, we have to integrate the velocity function V of TDT. So we have to perform two steps. First of all, let's identify the velocity function V of T, which is the integral of 3 + 2 cosine C. DT We're basically integrating our acceleration function. We can evaluate two integrals, right? First of all, the integral of 3DT is simply 3T and the integral of cosine is signed. So we're basically adding 2. S T plus a constant of integration C1. Now we can identify C1 from the initial condition V of 0 equals -1, meaning if T is 0, V is -1. So let's set V equals -1 and C equals 0. So we get -1 equals 3 multiplied by 0 + 2 sign of 0 plus C1. And from here we can show that C1 is equal to -1 because sin of 0 is 03 multiplied by 0 is 0, so we get -1 equals C1. Therefore, our velocity function V of T is 3T plus 2 T minus 1. And now we can focus on the position function as of T, which is the integral. Of the latter function 3 C plus. To sign. -1 DT Now we have 3 integrals to evaluate, right, because we have a sum and Difference We can factor out the first constant, which is 3, and integrate C, which is T2 divided by 2 according to the power rule. For the second integral, we have 2 sine C, so that'd be 2, we can factor out the constant, and the integral of sine is going to be negative cosine. So we have 2 multiplied by a negative cosine of T. Minus the integral of 1 dt, which is simply the integral of DT or Basically, it is equal to. T And we are adding a constant of integration C2. So now let's simplify. We got 3 halves T squared, minus 2 cosine of T minus T plus C2. And now we have our next condition as of 0 is 5, meaning. If T is 0, S is 5, we're settings equals 5. And for every T we're going to substitute 0, so we get 3 halves multiplied by 0 squared, minus 2 cosine of 0, minus 0 plus C2. Now, cosine of 0 is equal to. One And therefore we get 5 equals 0 minus 2 multiplied by 1, that's -2. Plus C2. And therefore, C2 is going to be equal to 5 + 2, which is 7. And now we have successfully identified the position function as of T, which is 3 halves. T squared Minus To cosine C. Minus T. Plus C2, so plus 7. And that will be our final answer for this problem. Thank you for watching.

Acceleration to position Given the following acceleration functions of an object moving along a line, find the position function with the given initial velocity and position.

a(t) = 2 + 3 sin t; v(0) = 1, s(0) = 10

Key Concepts

Acceleration Function

Integration

Initial Conditions

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