Question

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

object when it changes direction

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Answer 1

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

x(t) = 4t ² - 41t + 78

→ v(t) = dx(t)/dt = 8t - 41

Solve v(t) = 0:

8t - 41 = 0

8t = 41

t = 41/8 = 5.125

Just to confirm that the velocity indeed changes sign:

• Pick any time before this one to check the sign of v :

v (0) = 8•0 - 41 = -41 < 0

• Pick any time after and check the sign again:

v (6) = 8•6 - 41 = 7 > 0

Now just find the position at this time:

x (5.125) = -433/16 = -27.0625

which means the object is 27.0625 units on the negative x-axis.

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

4t ² - 41t + 78 = 4 (t ² - 41/4 t ) + 78

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

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

… = 4 (t - 41/8)² - 433/16

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