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An object moves according to a law of motion, where, its position is described by the following function, s = f(t) = t^4 - 4t + 1. The time t is measured in seconds and s in meter.a. Sketch the velocity graph and determine when is the object moving in the positivedirection.b. Draw a diagram of the motion of the object and determine the total distancetraveled during the first 6 seconds.

An object moves according to a law of motion where its position is described by the following function s ft t4 4t 1 The time t is measured in seconds and s in m class=

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SOLUTION

(a) The function given is the distance graph. The velocity fuction is the derivative of the distance function. This becomes

[tex]\begin{gathered} f(t)=t^4-4t+1 \\ f^{\prime}(t)=4t^3-4 \\ v=4t^3-4 \end{gathered}[/tex]

the graph is shown below

From the velocity graph above, the function is moving in a positive direction from 1 second and beyond

(b) The diagram of motion of the object is shown in the distance graph below

The total distance travelled is the integral of the velocity function from 0 to 6 as shown below

[tex]\begin{gathered} \int_0^64x^3-4 \\ =[\frac{4x^4}{4}-4x]_0^6 \\ =[x^4-4x]_0^6=6^4-4(6)-(0^4-4(0)) \\ =1296-24-0 \\ =1272 \end{gathered}[/tex]

Hence the answer is 1272 m

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