AP Samples – Distance and Displacement
1983 AB5, BC3
At time t = 0, a jogger is running at a velocity of 300 meters per minute. The jogger is slowing
down with a negative acceleration that is directly proportional to time t. This brings the jogger to
a stop in 10 minutes.
Use a (t )  kt
a) Write an expression for the velocity of the jogger at time t.
b) What is the total distance traveled by the jogger in that 10-minute interval?
1987 AB1 (part d)
A particle moves along the x-axis so that its acceleration at any time t is given by a(t )  6t  18 .
At time t = 0 the velocity of the particle is v(0)  24 , and at time t = 1 its position is x(1)  20
d) Find the total distance traveled by the particle from t = 1 to t = 3.
AP Samples – Distance and Displacement
1990 AB1 (part c)
A particle, initially at rest, moves along the x-axis so that its acceleration at any time t ≥ 0 (in
seconds) is given by a(t )  12t 2  4 . The position of the particle when t = 1 is x(1)  3 meters
c) Find the total distance traveled by the particle from t = 0 to t = 2.
(
)
v(t) = ò 12t 2 - 4 dt = 4t 3 - 4t + C
Since the particle is initially at rest, v(0)=0, so C=0
ò
2
0
4t 3 - 4t dt = 10
BATTAGLIA ORIGINALS
Using the information from 1990 AB1 (above) answer the following
B1) At what time, t, would the particle have a displacement of 63 meters?
B2) Give the particle’s position at the time found in B1 n.
B3) Give the intervals when the particle is speeding up and slowing down.
B1)
ò ( 4t
b
0
3
- 4t )dt = 63
b
t 4 - 2t 2 ùû 0 = 63
b4-2b2=63
b=3 seconds
AP Samples – Distance and Displacement
B2)
x(t) = ò ( 4t 3 - 4t )dt = t 4 - 2t 2 + C
x(1) = 3 = 1- 2 + C
C=4
x(t) = t 4 - 2t 2 + 4
x(3) = 34 - 2(3)2 + 4 = 67meters
B3)
v(t) = 4t 3 - 4t is positive when t > 1 and negative when 0 < t < 1
a(t) = 12t 2 - 4 is positive when t > .577 and negative when 0 < t < 0.577
The particle is speeding up from 0 < t < 0.577 and when t > 1.
The particle is slowing down when 0.577 < t < 1.