Thread: Calculus Question - position of particle speed problem.

This problem is taken off college board AP* sample exam.
An image of the problem is posted here for convenience.



Using speed instead of velocity is throwing me off, could you please do this step by step?

Thank you for your assistance!


There are also 5 possible choices, the correct answer is E. Also please note a calculator is not allowed on this section, so could you please explain this using techniques that can be done by hand only?

*AP and Advanced Placement Program are registered trademarks of the College Entrance Examination Board, which was not involved in the production of and does not endorse this product.

Originally Posted by whatsntomake

This problem is taken off collegeboard sample exam.
An image of the problem is posted here for convience.



Using speed instead of velocity is throwing me off, could you please do this step by step?

Thank you for your assistance!


There are also 5 possible choices, the correct answer is E. Also please note a calculator is not allowed on this section, so could you please explain this using techniques that can be done by hand only?

The "position" of a particle is a code name for the "displacement" of a particle, which is the vector version of distance from the origin. So when you take the derivative of the position function you get the velocity of the particle.

To get the speed of the particle at any time, simply take the magnitude of the velocity vector. In 1 - D motion this corresponds to taking the absolute value of the velocity.

So:
s(t) = 2t^3 - 24t^2 + 90t + 7

v(t) = 6t^2 - 48t + 90 <-- This is the velocity.

We wish to know for what times |v(t)| is increasing. For this we need the derivative of |v(t)| = |6t^2 - 48t + 90|

By the chain rule:
d|v(t)|/dt = sign(6t^2 - 48t + 90)*(12t - 48)

When is this positive?

We need to know when sign(6t^2 - 48t + 90) > 0
The zeros of 6t^2 - 48t + 90 are:
6t^2 - 48t + 90 = 0

t^2 - 8t + 15 = 0

(t - 3)(t - 5) = 0

t = 3 s, 5 s

So for t in the domain
[0, 3), 6t^2 - 48t + 90 > 0
(3, 5), 6t^2 - 48t + 90 < 0
(5, infinity), 6t^2 - 48t + 90 > 0

Thus sign(6t^2 - 48t + 90) > 0 for t belonging to [0, 3) and (5, infinity) and sign(6t^2 - 48t + 90) < 0 for t belonging to (3, 5).

Now, 12t - 48 is positive for t >4.

Putting it all together we get that:
d|v(t)|/dt > 0 for t belonging to (3,4) since both sign() and 12t - 48 are both negative. For the time period t > 5 s, both sign() and 12t - 48 are positive, so the speed is also increasing for t > 5 s.

Thus the speed is increasing for the time period 3 s to 4 s and t > 5 s, which is answer E.

This can all be done without a calculator (I didn't use one) but it IS a long question to answer.

-Dan