Topic 2.1 Extension 07
Acceleration
To follow Worksheet "Acceleration"
1)
Recall that the speed or velocity of an object is the time rate
of change of the position of the object, or in formula form <v> = x/t
and v = dx/dt. What is the difference between <v> and v?
<v> is average speed (macroscopic time interval)
v is instantaneous (microscopic time interval)
FYI: We define the average acceleration <a> of an object as the time
rate of change of the speed or velocity of the object, or in formula
form
<a> = v .
t
2)
Suppose the velocity of a particle is given by v = 10t – 5t2,
where v is in m/s and t is in seconds. Complete the table below:
t0(s) t1(s) v0(m/s)
v1(m/s)
v(m/s)
t(s) <a>(m/s2)
2.50
3.00
-6.25
-15.0000
-8.7500
0.50
-17.50
2.50
2.75
-6.25
-10.3125
-4.0625
0.25
-16.25
2.50
2.60
-6.25
-7.8000
-1.5500
0.10
-15.50
2.50
2.55
-6.25
-7.0125
-0.7625
0.05
-15.25
2.50
2.51
-6.25
-6.4005
-0.1505
0.01
-15.05
FYI: We define the instantaneous acceleration a as the limit as t
approaches zero of the average acceleration. Thus
a = lim
t0
v
t
=
dv .
dt
FYI: Of course, since we’ll be working in the x, y and z directions,
we actually have three instantaneous accelerations defined:
ax =
dvx
,
dt
ay =
dvy
,
dt
az =
dvz
.
dt
3)
Which of one the above values of <a> is closest to the
instantaneous acceleration of the particle at t = 2.50 s?
The last one has the smallest time interval (0.01 seconds) and so is
the best. a = -15.05 m/s/s.
4)
Now using the formula v = 10t – 5t2 and the techniques learned in
the previous lessons, find the instantaneous acceleration of this
10 – 10t
particle at any time t (a = dv/dt). a(t) = _________________.
What
–15 m/s2
is a(2.50)? ___________.
How does this compare with your answer from
3?
Very close (compare to -15.05 m/s2).
5)
Perhaps you recall from kinematics the equation
1
y(t) = yo + vot +
at2,
2
where y is the position or height of a particle at time t, yo is the
original height (a constant), vo is the original speed (a constant) and
a is the acceleration (another constant).
a)
Find v(t) = dy/dt.
v = v0 + at
1
b)
Does this formula look familiar?
Yes
c)
Now find a = dv/dt for the function you obtained in a).
a = a
d)
Yes
Is the acceleration of the particle constant? _____.
do you know?
Because there is no time dependence (no t).
How
6)
For practice, consider a particle whose y-position is given by
y(t) = 250 + 50t - 10t2 where t is in seconds and y is in meters.
a)
Find the speed of the particle at any time t.
v = dy/dt = 50 – 20t
b)
Find the speed of the particle at t = 10 s.
v(10) = 50 – 20(10) = -150 m/s
c)
Find the acceleration of the particle at any time t.
a = dv/dt = -20
d)
Find the acceleration of the particle at t = 10 s.
a = -20 m/s2
e)
Is the acceleration of the particle constant?
Yes
f)
What is the average speed of the particle for the time
interval 0  t  10 s?
<v> = v = v(10) - v(0) = 50-20(10) – [50-20(0)] = -20 m/s
10 - 0
10
t
7)
For further practice, consider a particle whose y-position is
given by y(t) = 250 + 5t - 10t2 + 15t3 where t is in seconds and y is in
meters.
a)
Find the speed of the particle at any time t.
v = dy/dt = 5 – 20t + 45t2
b)
Find the speed of the particle at t = 10 s.
v(10) = 5 – 20(10) + 45(10)2 = 4305 m/s
c)
Find the acceleration of the particle at any time t.
a = dv/dt = -20 + 90t
d)
Find the acceleration of the particle at t = 10 s.
a(10) = -20 + 90(10) = 880 m/s2
e)
Is the acceleration of the particle constant?
No, it depends on time: a = -20 + 90t.
f)
What is the average speed of the particle for the time
interval 0  t  10 s?
2
<v> = v = v(10) - v(0) = 4305 – [5-20(0)+45(0) ] = 430 m/s
10 - 0
10
t
2