The velocity function (in meters per second) for a particle moving along a line is given by
v(t)=3t−5,0≤t≤3.
(a) Find the displacement (in meters) of the particle.
Displacement = meters
(b) Find the total distance traveled (in meters) by the particle.
Total distance traveled = meters
a)
Velocity: v(t) = 3t − 5 , 0 < t < 3
Take x(0) = 0. Then the displacement is
t
t
0
0
x(t) = ∫ v(t ′ ) dt ′ = ∫ (3t ′ − 5) dt ′ =
The position at t = 3 is x(3) = −
3 2
1
t − 5t = 3t − 10 t
2
2
(
)
3
m
2
b)
Particle moves to left initially, reaches some leftmost point xmin , then reverses
and moves to right.
If we consider an "odomoter"-like measure of total distance,
that is, irrespective of direction, then the total distance travelled is
(
)
d = −xmin + x(3) − xmin = x(3) − 2xmin
Evaluating,
(−5)
5
tmin = −
=
2(3 2) 3
xmin = x(5 3) =
x(3) = −
⎞5
1⎛ 5
25
3
−
10
=
−
⎟⎠ 3
2 ⎜⎝ 3
6
3
2
3 25 41
d=− +
=
2 3
6
However, if we want instead to find just the overall change in position,
3
then d =
to the left.
2
The acceleration function (in m/s2) and initial velocity for a particle moving along a line
is given by
a(t)=−9t−36,v(0)=−45,0≤t≤10.
(a) Find the velocity (in m/s) of the particle at time t.
Velocity = meters
(b) Find the total distance traveled (in meters) by the particle.
Total distance traveled = meters
a)
a(t) = −9t − 36 = −9(t + 4),
v(0) = −45,
0 ≤ t ≤ 10
t
⎛1
⎞
9
v(t) = v(0) − 9 ∫ (t ′ + 4) dt ′ = −45 − 9 ⎜ t 2 + 4t ⎟ = − t 2 − 36t − 45
2
⎝2
⎠
0
(
)
9 2
t + 8t + 10
2
The velocity at t = 10s is
9
v(10) = − 190 = −855 m/s
2
=−
( )
b)
In this case, the particle moves monotonically to the left, so there is no ambiguity, and the
total distance travelled is
10
9 10
d = −x(10) = ∫ v(t ′ ) dt ′ = ∫ (t ′ 2 + 8t ′ + 10) dt ′
20
0
=
⎤
9 ⎡1
3
2
⎢ (10) + 4(10) + 10(10) ⎥ = 3,750 m
2 ⎣3
⎦
The marginal cost of manufacturing x yards of a certain fabric is
C′(x)=3−0.01x+0.000006x2 (in dollars per yard). Find the increase in cost if the
production level is raised from 2500 to 7500 yards.
Increase in cost = $
C ′(x) = 3 − 0.01x + 0.000006x 2 $/yd
C(x) = 3x − 0.01x 2 + 0.000006x 3
$
So the increase in cost at 7500 yards compared to 2500 yards is
I = C(7500) − C(2500)
= 3(7500 − 2500) − 0.01(75002 − 25002 ) + 0.000006(75003 − 25003 )
= $1,952,500