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puteri [66]
2 years ago
11

An object moves along the x-axis. Its position is given by the equation y( X = 4t^2 - 41t+ 78 \). Find the position of the

Physics
1 answer:
prohojiy [21]2 years ago
7 0

The object changes direction when its velocity changes sign. You can get the velocity function by differentiating the position function with respect to time <em>t</em> :

<em>x(t)</em> = 4<em>t</em> ² - 41<em>t</em> + 78

→   <em>v(t)</em> = d<em>x(t)</em>/d<em>t</em> = 8<em>t</em> - 41

Solve <em>v(t)</em> = 0:

8<em>t</em> - 41 = 0

8<em>t</em> = 41

<em>t</em> = 41/8 = 5.125

Just to confirm that the velocity indeed changes sign:

• Pick any time before this one to check the sign of <em>v</em> :

<em>v</em> (0) = 8•0 - 41 = -41 < 0

• Pick any time after and check the sign again:

<em>v</em> (6) = 8•6 - 41 = 7 > 0

Now just find the position at this time:

<em>x</em> (5.125) = -433/16 = -27.0625

which means the object is 27.0625 units on the negative <em>x</em>-axis.

You can also do this without calculus by completing the square in the position function:

4<em>t</em> ² - 41<em>t</em> + 78 = 4 (<em>t</em> ² - 41/4 <em>t</em> ) + 78

… = 4 (<em>t</em> ² - 2• 41/8 <em>t</em> + (41/8)² - (41/8)²) + 78

… = 4 (<em>t</em> ² - 2• 41/8 <em>t</em> + (41/8)²) - 4•1681/64 + 78

… = 4 (<em>t</em> - 41/8)² - 433/16

which describes a parabola that opens upward. When <em>t</em> = 41/8 = 5.125, the quadratic term vanishes and the turning point of the parabola occurs at a position of -433/16 units.

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