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89

Step-by-step Answer
PAGE: 189SET: ExercisesPROBLEM: 89
Please look in your text book for this problem Statement

(a)

Motion of particle is .

Find the velocity of particle by applying derivative.

Therefore the velocity of particle is .

Equate it to zero.

So the critical number is .

(b)

Consider the test intervals to find the interval of increasing and decreasing.

Test interval Test value Sign of  Conclusion

Increasing

Decreasing

 

The sign of the velocity function is positive on the intervals .

(c)

The sign of the velocity function is negative on the intervals .

(d)

At , the velocity function changes its sign.

(a) The velocity of particle is .

(b) The sign of the velocity function is positive on the intervals .

(c) The sign of the velocity function is negative on the intervals .

(d) Velocity function changes its direction at .



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