Motion Along a Coordinate Line

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Question

Motion Along a Coordinate Line

Exercises 1–6 give the positions s = f(t) of a body moving on a coordinate line, with s in meters and t in seconds.

b. Find the body’s speed and acceleration at the endpoints of the interval.

s = 25/t² − 5/t, 1 ≤ t ≤ 5

Accepted Answer

Welcome back, everyone. For a particle moving in a straight line, as position is given by S equals 9 divided by T2 minus V divided by T, where S is in meters and T is in seconds. Calculate the particle's velocity and acceleration at T equals 3. What we're going to do is basically recall that velocity as a function is the first derivative of the position functions we have to differentiate our position function, and then because we're also looking at acceleration, while it is the second derivative of position, we're basically the first derivative of velocity. So essentially we can begin with finding the velocity function. B of T is going to be the derivative of S of T, right? So we want to differentiate 9 divided by T2 minus 3 divided by T. And we can do that using the power rule if we rewrite our terms using the reciprocal rules, where differentiating 9, T raise the power of -2 minus 3, T raise the power of -1. Let's identify the derivative. So basically the derivative of T raises the power of -2, according to the power rule is -2, multiplied by T raise to the power of -3, so we get 9 multiplied by -2, T to the power of -3, and then we are subtracting 3 multiplied by -1. T raises the power of -1 minus 1, which is -2. Let's simplify. We're going to get -18, T raise the power of -3, plus 3T raise the power of -2, or basically, 3 Divided by T2 minus 18 divided by TQ. We want to identify the particle's velocity at time of 3, right? So let's evaluate V. Actually, B, right, because we're looking for velocity, so B at 3. This is going to give us 3 divided by T2d. So essentially we're taking 3 squared and subtracting 18 divided by 3 cubed. Let's find the common denominator. It is going to be 3 cubed, so we get 3 multiplied by 3 minus 18 divided by the common denominator of 3 cubed, which is 27. So now 9 minus 18 gives us -9 and -9 divided by 27 is -13. And the units are meters per second, right? So we have got the velocity. Let's label our answer, and now let's continue with the acceleration function. A of T is the first derivative of velocity. Let's differentiate our expression using the power rule once again. -18, multiplied by the derivative of T, raises the power of -3. Plus 3 multiplied by the derivative of T raises the power of -2. According to the power rule, we're going to get -1 multiplied by -3 T to the power of -4. Plus 3 multiplied by -2, T raises the power of -3. Now, let's rewrite it in a more readable form, so -18 multiplied by -3 is 54. We're dividing that by T to the power of 4. And then we are subtracting 6 divided by T cube. Now, acceleration at time 3 is going to be equal to 54. Divided by 3 to the power of 4. Minus 6. Divided by 3 to the power of 3. Once again, we're going to find a common denominator, we end up with 54 minus 6 multiplied by 3, divided by 3 to the power of 4, which is 81. So now what do we get? 54 minus 18 is 36, right? So we can say this is 36 divided by 81, and we can simplify. They're both divisible by 9, so we get 4 divided by 9 m per second squared, right? Those are the units for acceleration. That would be our second answer, and thank you for watching.

Motion Along a Coordinate Line

Exercises 1–6 give the positions s = f(t) of a body moving on a coordinate line, with s in meters and t in seconds.

b. Find the body’s speed and acceleration at the endpoints of the interval.

s = 25/t² − 5/t, 1 ≤ t ≤ 5

Key Concepts

Derivative

Velocity and Speed

Second Derivative and Acceleration

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