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Chapter 2: Problem 49

An object's position as a function of time \(t\) is given by \(x=b t^{4}\) with \(b\) a constant. Find an expression for the instantaneous velocity, and show that the average velocity over the interval from \(t=0\) to any time \(t\) is one- fourth of the instantaneous velocity at \(t\)

Short Answer

Expert verified
The instantaneous velocity of the object is \(4bt^3\) while the average velocity over the interval from \(t=0\) to any time \(t\) is \(bt^3\). Indeed, the average velocity is one-fourth of the instantaneous velocity at time \(t\).

Step by step solution

01

Calculating the Instantaneous Velocity

The instantaneous velocity can be obtained by taking the derivative of the position function with respect to time. Hence, the velocity \(v(t)\) equals to the derivative of \(x(t)\). This gives: \[v(t) = \frac{dx}{dt}=\frac{d}{dt}(bt^4)\] After applying the power rule for derivatives, the instantaneous velocity becomes: \[v(t)=4bt^3\]
02

Finding the Average Velocity

The average velocity is determined by dividing the total displacement (which is the difference in the position from \(t=0\) to any time \(t\)) by the time taken. Here, the average velocity \(V_{avg}\) is given by: \[V_{avg}=\frac{x(t) - x(0)}{t-0}\] Since the function \(x(t)=bt^4\), then at \(t=0,\) \(x(0)=0\). So the average velocity becomes: \[V_{avg}= \frac{bt^4-0}{t} = bt^3\]
03

Proving the relation between the Instantaneous and Average Velocities

In the final step, it should be shown that the average velocity is one-fourth of the instantaneous velocity. With the instantaneous velocity \(v(t)=4bt^3\) and the average velocity \(V_{avg}=bt^3\), when we divide the instantaneous velocity by the average velocity, we get: \[\frac{v(t)}{V_{avg}}=\frac{4bt^3}{bt^3}=4\] Therefore, the average velocity is indeed one-fourth of the instantaneous velocity at time \(t\).

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