Examples of Motion Problems
In solving motion problems, you should always
remember to start from the basics.
For the discrete case, these are:
• vx-average = Dx /Dt and
• ax-average = Dvx /Dt . .
For the continuous case when we have
constant acceleration, these become:
• v(t) = vo + a*t and
• s(t) = so + vo*t + ½ a*t2 (where s can be either x or y
depending on whether the motion is horizontal or
vertical).
Example: Discrete Case
Given the following values of velocity at the
indicated times, find the average acceleration
between t = 4 seconds and t = 5 seconds, and
find the position at t = 5 seconds given that the
initial position is -7 m.
t (s) v (m/s)
t (s)
v (m/s)
0 - 4.00
3
+ 7.95
1 - 1.95
4
+10.00
2 + 3.00
5
+ 7.95
Example: Discrete Case
Diagram:
x=0
t (s)
0
1
2
v (m/s)
- 4.00
- 1.95
+ 3.00
t (s)
3
4
5
v (m/s)
+ 7.95
+10.00
+ 7.95
Example: Discrete Case
Since we are working with discrete values, we start
with the definitions of average velocity and
average acceleration:
ax-average = Dvx /Dt and vx-average = Dx /Dt
For the average acceleration between t = 4 s and t
= 5 s, we can apply the average acceleration
formula directly:
ax-average = Dvx /Dt becomes
aavg = Dvx /Dt = (7.95 m/s – 10 m/s) / (5s – 4s)
= -2.05 m/s2 .
Example: Discrete Case
To find the position at t = 5 sec., we must
start at the one position we know, which is
x(t=0s) = xo = -7 m. and use the
definition of average velocity:
vx-average = Dx /Dt , or Dx = vx-average * Dt
or Dx = xf – xi = v(between tf and ti)*(tf-ti)
However, because the velocity changes, we
need to do this in individual steps:
Example: Discrete Case
Dx = xf – xi = v(between tf and ti)*(tf-ti) , or
xf = xi + v(between tf and ti)*(tf-ti)
x1 = x0 + v(between t1 and t0)*(t1-t0)
x1 = -7 m + ½(-4.0 m/s + -1.95 m/s) * (1s – 0s)
= -7 m + (-2.98 m/s)*(1 s) = -9.98 m = x1
x2 = x1 + v(between t2 and t1)*(t2-t1)
x2 = -9.98 m + ½(-1.95 m/s + 3.0 m/s) * (2s – 1s)
= -9.98 m + (.53 m/s)*(1 s) = -9.46 m = x2
x3 = -9.46 m + ½(3.0 m/s + 7.95 m/s) * (3s – 2s)
= -9.46 m + (5.48 m) = -3.98 m = x3
Example: Discrete Case
x4 =
=
x5 =
=
-3.98 m + ½(7.95 m/s + 10.0 m/s) * (4s – 3s)
-3.98 m + (+8.98 m/s)*(1 s) = 5.00 m = x4
5.00 m + ½(10.0 m/s + 7.95 m/s) * (5s – 4s)
5.00 m + (+8.98 m) = +13.98 m = x5
In this particular case, the velocity data was obtained from the
functional form:
v(t) = +3 m/s – (7 m/s)*cos({p rad/4 sec}*t)
[note that {p rad / 4 sec} = {45o/sec} ]
Which leads (via calculus) to functional forms for x(t) and a(t) of:
a(t) = (7*p m/ 4 s2)*sin({p rad/4 sec}*t)
x(t) = -7 m + (3 m/s)*t – (28 m/p)* sin({p rad/4 sec}*t)
Example: Discrete Case
a(t) = (7*p m/ 4 s2)*sin({p rad/4 sec}*t)
x(t) = -7 m + (3 m/s)*t – (28 m/p)* sin({p rad/4 sec}*t)
If we use the above functional forms to calculate the
acceleration between t = 4 sec and t = 5 sec, that is, at t
= 4.5 sec, we get:
a(t) = (7*p m/ 4 s2)*sin({p rad/4 sec}*4.5 sec) = -2.10 m/s2
which compares to our numerical result of -2.05 m/s2.
If we use the above functional forms to calculate the
position at t = 5 sec, we get: x(5 sec) =
-7 m + (3 m/s)*(5 s) – (28 m/p)* sin({p rad/4 sec}*(5 s)) =
-7 m + 15 m + 6.30 m = 14.30 m which compares to our
numerical result of 13.98 m.
Example: Discrete Case
a (functional result) = -2.10 m/s2
a (numerical result) = -2.05 m/s2
x (functional result) = 14.30 m
x (numerical result) = 13.98 m
Note that the functional and numerical results are
close but not exactly the same. The numerical
result is based on discrete data which is
incomplete, whereas the functional result is
based on complete (continuous) data.
Example: plane taking off
A plane takes off from an aircraft carrier. Assume
the following:
• the plane accelerates with a constant
acceleration,
• the flight deck is 80 meters long,
• the initial speed of the plane is zero,
• the final speed of the plane is 85 m/s.
What is the acceleration of the plane during takeoff, and how long a time does the take-off take?
Plane taking off – continued
Draw a diagram:
to = 0 s
vo = 0 m/s
tf = ?
a=?
Dx = 80 meters
vf = 85 m/s
Plane taking off – continued
Since this is a constant acceleration problem
along the horizontal, we have two equations:
• v(t) = vo + a*t and
• x(t) = xo + vo*t + ½ a*t2 .
We are given the following:
vo = 0 m/s; vf = 85 m/s; xf – xo = 80 meters
(from this we can let xo = 0 m, so xf = 80 m).
We are asked for the acceleration, a, and the
time, t.
Plane taking off – continued
Since we have two equations and two
unknowns (a and t), we should be able to
solve for both a and t.
We now substitute the knowns into our two
equations:
v(t) = vo + a*t becomes 85 m/s = 0 m/s + a*t
and
x(t) = xo + vo*t + ½ a*t2 becomes
80 m = 0 m + (0 m/s)*t + ½ a*t2 .
Plane taking off – continued
85 m/s = 0 m/s + a*t
80 m = 0 m + (0 m/s)*t + ½ a*t2
Note that both equations have both
unknowns, so we have to use the
techniques of simultaneous equations.
If we solve for a in the first equation, we get:
a = 85 m/s / t , and now we use this in the
second equation:
80 m = 0 m + (0 m/s)*t + ½ (85 m/s / t)*t2.
Plane taking off – continued
80 m = 0 m + (0 m/s)*t + ½ (85 m/s / t)*t2
This is now one equation in one unknown (t)
80 m = (42.5 m/s)*t , or
t = (80 m) / (42.5 m/s) = 1.88 seconds.
We can now use this value of t in our first
equation: 85 m/s = 0 m/s + a*t to get:
85 m/s = 0 m/s + a*(1.88 sec), or
a = (85 m/s) / (1.88 sec) = 45.16 m/s2 .
Plane taking off – continued
t = 1.88 seconds.
a = 45.16 m/s2 .
Are these results reasonable?
A plane taking off from an aircraft carrier on a short flight
deck certainly has to take off in a short time, so about 2
seconds looks reasonable.
The acceleration of 45.16 m/s2 is equivalent to 4.61 gees’
(1 gee = 9.8 m/s2). This is a pretty big acceleration, but
that also sounds reasonable given the circumstances.
By the way, this type of acceleration will tend to make
most people black out – not a good thing when taking off
in a plane.
Example: Throwing an object up
A candy bar is thrown upwards from a
person on the ground to a person on a
balcony 15 meters above the ground. The
person on the ground throws the candy
bar up at a speed of 3 m/s. How long will
it take the candy bar to reach the person
on the balcony?
Example: Throwing an object up
y = 15 m
v=?
t=?
a = g = -9.8 m/s2
yo = 0 m
vo = 3 m/s
to = 0 s
We recognize this as
a “falling” problem,
so we can use the
equations for
constant acceleration
with the acceleration
equaling that of
gravity.
Example: Throwing an object up
v(t) = vo + a*t and
y(t) = yo + vo*t + ½ a*t2
where we identify the following:
y = 15 m (final position is 15 meters above the ground)
yo = 0 m (initial position is on the ground)
v=?
(speed when it reaches the person on the balcony)
vo = 3 m/s (initial speed)
t=?
(time when it reaches the person on the balcony)
a = g = -9.8 m/s2 (acceleration due to gravity)
Example: Throwing an object up
v(t) = vo + a*t and
y(t) = yo + vo*t + ½ a*t2
Substituting in the values we know:
v(t) = (3 m/s) + (-9.8 m/s2) *t and
15 m = 0 m + (3 m/s)*t + ½ (-9.8 m/s2)*t2
We see that we have two equations for two
unknowns, but the second equation has
only one unknown, t. However, it is a
quadratic equation in that unknown.
Example: Throwing an object up
v(t) = (3 m/s) + (-9.8 m/s2) *t and
15 m = 0 m + (3 m/s)*t + ½ (-9.8 m/s2)*t2
To solve quadratic equations, we need to put this
in standard form,
a*x2 + b*x + c = 0
½ (-9.8 m/s2)*t2 + (3 m/s)*t - 15 m = 0 m
and then use the quadratic formula
x = [-b ± {b2 – 4*a*c}1/2 ] / [2*a]
Where a = -4.9 m/s2, b = 3 m/s, and c = -15 m.
Example: Throwing an object up
x = [-b ± {b2 – 4*a*c}1/2 ] / [2*a]
where a = -4.9 m/s2, b = 3 m/s, and c = -15 m.
t = [-(3 m/s) ± {(3 m/s)2 – 4*(-4.9 m/s2)*(-15 m)}1/2 ] / [2*(-4.9 m/s2)]
=
[-(3 m/s) ± {-285 m2/s2 }1/2 ] / [-9.8 m/s2] .
Note that we have the square root of a negative
number in the formula. This means that the
math gives us an imaginary number for the time.
What does this mean physically?
Example: Throwing an object up
t = [-(3 m/s) ± {-285 m2/s2 }1/2 ] / [-9.8 m/s2] .
What does this mean physically?
You may have noticed that the person on the
ground did not throw the candy bar up with a
very big speed, only 3 m/s. What happens is
that the candy bar never makes it up to 15
meters. We only “imagine” that the candy bar
gets up that high. The math is trying to tell us
that the candy bar never reaches that high.
So what do we have to do to get the candy bar
that high?
Example: Throwing an object up
x = [-b ± {b2 – 4*a*c}1/2 ] / [2*a]
where a = -4.9 m/s2, b = 3 m/s, and c = -15 m.
We see from the formula that to make x real, the
quantity b2 has to be greater than (4*a*c).
In our case, to make t real, the initial velocity (which
is the “b”) must be at least as big as the quantity
[ 4*(½g)*(y)] (which is the “4*a*c”).
Therefore, let’s retry this problem with a bigger
initial speed, say 20 m/s.
Example: Throwing an object up
x = [-b ± {b2 – 4*a*c}1/2 ] / [2*a]
where a = -4.9 m/s2, b = 20 m/s, and c = -15 m.
t = [-(20 m/s) ± {(20 m/s)2 – 4*(-4.9 m/s2)*(-15 m)}1/2 ] / [2*(-4.9 m/s2)]
=
[-(20 m/s) ± {106 m2/s2 }1/2 ] / [-9.8 m/s2] .
What does the plus or minus sign in the formula
mean? It actually gives us two answers:
t = 0.99 seconds, and t = 3.09 seconds.
If we look at our diagram, we see that the candy bar
passes the person on the way up (at t = 0.99 s) and
on the way back down (at t = 3.09 s).
Example: Throwing an object up
Now that we know the times when the candy
bar reaches the person on the balcony, we
could use the other (velocity) equation to
find the speeds at these two times.
v(t) = (20 m/s) + (-9.8 m/s2) *t
On the way up, t = 0.99 s, so
v(up)
= (20 m/s) + (-9.8 m/s2) *(0.99 s) = +10.3 m/s,
and on the way down, t = 3.09 s, so
v(down) = (20 m/s) + (-9.8 m/s2) *(3.09 s) = -10.3 m/s .