AP PHYSICS C
Kinematics and Derivatives
NAME _______________________
1. Find the velocity and acceleration functions for the objects that obey the following position functions.
a. x = 9 + 3t + 2t2– 4t3
b. x = 5t2 - 8t – 15
c. x = 7t4 – 8t2
d. x = 8t - 7t2
2. Find the time(s) at which the objects in #1 have the following characteristics (Use the function with the
matching letter).
a. velocity is zero
b. moving in the positive direction (+ velocity)
c. has a position in the positive x area
d. has a negative acceleration
3. OMIT Find the position at t = 4 seconds for the objects with the following velocity or acceleration
functions. Assume the initial positions and initial velocities are zero unless the equations given
specifically tell you otherwise.
a. v = 24 – 5t
b. a = -3 t + 5t2
4. OMIT Give a description of what could be causing the motion described in #3 a and b.
1. Use the power rule!
a. 𝑣 = −12𝑡 2 + 4𝑡 + 3
𝑎 = −24𝑡 + 4
b. 𝑣 = 10𝑡 − 8
𝑎 = 10
c. 𝑣 = 28𝑡 3 − 16𝑡
𝑎 = 84𝑡 2 − 16
d. 𝑣 = −14𝑡 + 8
𝑎 = −14
2.
a. Set v = 0 in the velocity equation you derived. The equation is quadratic, so try solving using the
quadratic formula to solve for t. You will have two answers, but only one is positive and real. T =
0.69s
b. You may notice the velocity equation you derived for b is a simple straight line with a slope and
intercept. The coefficient in front of t is the slope of the velocity curve. The velocity is positive
when the velocity curve is in the positive half of the graph. Solve for t when v = 0 and once you
have a value, you can test whether the velocity is positive or negative in the time before or after your
solved t. However, testing values is not necessary in this case since we know the slope is positive,
therefore the time after t when v=0 is always going to have a positive velocity. Velocity is always
positive when t > 0.8s.
c. Factor out t2 from the position equation and notice the root terms, (t2) and (7t2 – 8). Since these
terms are multiplied together, at any time where either term is equal to zero, the whole equation will
equal zero. Separate the two terms and set them equal to zero to solve for the times in which the
equation will equal zero. Answers are t = 0, sqrt(8/7) seconds. We have solved for the times where
the position is equal to zero, but to solve for when the position in in the positive area, one can test for
values just before or after these t values. Positive position after t = sqrt(8/7) s.
d. The acceleration equation for d is just a constant, therefore acceleration does not change over time.
It is constant at -14 and therefore always has a negative acceleration