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is moving with respect to the stationary ground; we can then find the velocity of the plane or boat with respect to the ground from the vector sum in Eq.
Typology: Summaries
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In three dimensions, the location of a particle is specified by its location vector, r:
r = xi + yj + zk (3.1)
If during a time interval ∆t the position vector of the particle changes from r 1 to r 2 , the displacement ∆r for that time interval is
∆r = r 1 − r 2 (3.2) = (x 2 − x 1 )i + (y 2 − y 1 )j + (z 2 − z 1 )k (3.3)
If a particle moves through a displacement ∆r in a time interval ∆t then its average velocity for that interval is
v =
∆r ∆t
∆x ∆t
i +
∆y ∆t
j +
∆z ∆t
k (3.4)
As before, a more interesting quantity is the instantaneous velocity v, which is the limit of the average velocity when we shrink the time interval ∆t to zero. It is the time derivative of the position vector r:
v =
dr dt
d dt
(xi + yj + zk) (3.6)
dx dt
i +
dy dt
j +
dz dt
k (3.7)
can be written: v = vxi + vy j + vzk (3.8)
where
vx =
dx dt
vy =
dy dt
vz =
dz dt
The instantaneous velocity v of a particle is always tangent to the path of the particle.
If a particle’s velocity changes by ∆v in a time period ∆t, the average acceleration a for that period is
a =
∆v ∆t
∆vx ∆t
i +
∆vy ∆t
j +
∆vz ∆t
k (3.10)
but a much more interesting quantity is the result of shrinking the period ∆t to zero, which gives us the instantaneous acceleration, a. It is the time derivative of the velocity vector v:
a =
dv dt
d dt
(vxi + vy j + vzk) (3.12)
=
dvx dt
i +
dvy dt
j +
dvz dt
k (3.13)
which can be written: a = axi + ayj + az k (3.14)
where
ax =
dvx dt
d^2 x dt^2
ay =
dvy dt
d^2 y dt^2
az =
dvz dt
d^2 z dt^2
When the acceleration a (for motion in two dimensions) is constant we have two sets of equations to describe the x and y coordinates, each of which is similar to the equations in Chapter 2. (Eqs. 2.6—2.9.) In the following, motion of the particle begins at t = 0; the initial position of the particle is given by
r 0 = x 0 i + y 0 j
and its initial velocity is given by
v 0 = v 0 xi + v 0 yj
and the vector a = axi + ay j is constant.
vx = v 0 x + axt vy = v 0 y + ay t (3.16) x = x 0 + v 0 xt + 12 axt^2 y = y 0 + v 0 yt + 12 ay t^2 (3.17) v^2 x = v^20 x + 2ax(x − x 0 ) v y^2 = v^20 y + 2ay (y − y 0 ) (3.18) x = x 0 + 12 (v 0 x + vx)t y = y 0 + 12 (v 0 y + vy)t (3.19) Though the equations in each pair have the same form they are not identical because the components of r 0 , v 0 and a are not the same.
rBA = position of B’s origin, as measured by A
with v’s and a’s standing for the appropriate time derivatives, then we have the relations:
rP A = rP B + rBA (3.23)
vP A = vP B + vBA (3.24)
For the purposes of doing physics, it is important to consider reference frames which move at constant velocity with respect to one another; for these cases, vBA = 0 and then we find that point P has the same acceleration in these reference frames:
aP A = aP B
Newton’s Laws (next chapter!) apply to such a set of inertial reference frames. Observers in each of these frames agree on the value of a particle’s acceleration. Though the above rules for translation between reference frames seem very reasonable, it was the great achievement of Einstein with his theory of Special Relativity to understand the more subtle ways that we must relate measured quantities between reference frames. The trouble comes about because time (t) is not the same absolute quantity among the different frames.
Among other places, Eq. 3.24 is used in problems where an object like a plane or boat has a known velocity in the frame of (with respect to) a medium like air or water which itself is moving with respect to the stationary ground; we can then find the velocity of the plane or boat with respect to the ground from the vector sum in Eq. 3.24.
3.2 Worked Examples
(a) The velocity vector v is the time–derivative of the position vector r:
v =
dr dt
d dt
(3. 0 ti − 4. 0 t^2 j + 2. 0 k) = 3. 0 i − 8. 0 tj
where we mean that when t is in seconds, v is given in m s.
(b) At t = 2.00 s, the value of v is
v(t = 2.00 s) = 3. 0 i − (8.0)(2.0)j = 3. 0 i − 16 .j
that is, the velocity is (3. 0 i − 16 .j) m s.
(c) Using our answer from (b), at t = 2.00 s the magnitude of v is
v =
√ v x^2 + v y^2 + v z^2 =
√ (3. 00 m s )^2 + (− 16. m s )^2 + (0)^2 = 16. m s
we note that the velocity vector lies in the xy plane (even though this is a three–dimensional problem!) so that we can express its direction with a single angle, the usual angle θ measured anti-clockwise in the xy plane from the x axis. For this angle we get:
tan θ =
vy vx
= − 5. 33 =⇒ θ = tan−^1 (− 5 .33) = − 79 ◦^.
When we take the inverse tangent, we should always check and see if we have chosen the right quadrant for θ. In this case − 79 ◦^ is correct since vy is negative and vx is positive.
(a) To clarify matters, what we mean here is that when we use the numerical value of t in seconds, we will get the values of r in meters. Since the velocity vector is the time–derivative of the position vector r, we have:
v =
dr dt =
d dt
(i + 4t^2 j + tk) = 0 i + 8tj + k
That is, v = 8tj + k. Here, we mean that when we use the numerical value of t in seconds, we will get the value of v in m s.
(b) The acceleration a is the time–derivative of v, so using our result from part (a) we have:
a =
dv dt =
d dt
(8tj + k) = 8 j
then r is the anti-derivative of v. We get:
r =
t^2 +
t^5 + C 3
] i +
[ 2 t^2 −
t^4 + 50t + C 4
] j
and once again we need to solve for the constants. x = 0 at t = 0, so
35 2
and y = 0 at t = 0, so
and so r is fully determined:
r =
[ 35 2
t^2 +
t^5
] i +
[ 2 t^2 −
t^4 + 50t
] j
Now we can answer the questions. We want to know the value of r (the particle’s position) at t = 3.0 s. Just plug in!
x(t = 3.0 s) =
(3.0)^5 = 181 m
and
y(t = 3.0 s) = 2(3.0)^2 −
(3.0)^4 + 50(3.0) = 161 m.
The components of the velocity at t = 3.0 s are
vx(t = 3.0 s) = 35(3.0) + 12 (3.0)^4 = 146 m s
and vy (t = 3.0 s) = 4(3.0) − 13 (3.0)^3 + 50 = 53 m s.
Here we have been careful to include the proper (SI) units in the final answers because coordinates and velocities must have units.
(a) Since we are given that the acceleration is constant, we can use Eqs. 3.16:
vx = v 0 x + axt vy = v 0 y + ay t
to get:
ax =
vx − v 0 x t
(20. 0 m s − 4. 0 m s ) 20 .0 s
= 0. 80 m s 2
and
ay =
vy − v 0 y t
(− 5. 0 m s − 1. 0 m s ) 20 .0 s
= − 0. 30 m s 2
and the acceleration vector of the fish is
a = (0. 80 m s 2 )i − (0. 30 m s 2 )j.
(b) With the angle θ measured counterclockwise from the +x axis, the direction of the acceleration a is:
tan θ =
ay ax
A calculator gives us: θ = tan−^1 (− 0 .375) = − 20. 6 ◦
Since the y component of the acceleration is negative, this angle is in the proper quadrant. The direction of the acceleration is given by θ = − 20. 6 ◦. (The same as θ = 360◦^ − 20. 6 ◦^ =
(c) We can use Eq. 3.17 to find the values of x and y at t = 25 s:
x = x 0 + v 0 x + 12 axt^2 = 10 m + 4. 0 m s (25 s) + 12 (0. 80 m s 2 )(25 s)^2 = 360 m
and
y = y 0 + v 0 y + 12 ayt^2 = − 4 .0 m + 1. 0 m s (25 s) + 12 (− 0. 30 m s 2 )(25 s)^2 = − 72 .8 m
At t = 25 s the velocity components of the fish are given by:
vx = v 0 x + axt = 4. 0 m s + (0. 80 m s 2 )(25 s) = 24 m s
and
vy = v 0 y + ayt = 1. 0 m s + (− 0. 30 m s 2 )(25 s) = − 6. 5 m s
Our equation for the y coordinate is
y = y 0 + y 0 yt + 12 ayt^2 = 0 + 0 + 12 (−g)t^2 = − 12 gt^2
We can now ask: “At what time t does y equal − 1. 9 × 10 −^2 m?”. Substitute y = − 1. 9 × 10 −^2 m and solve:
t^2 = − 2 y g
2(− 1. 9 × 10 −^2 m)
= 3. 9 × 10 −^3 s^2
which gives:
t = 6. 2 × 10 −^2 s
Since this is the time of impact with the target, the time of flight of the bullet is t =
(b) The equation for x−motion is
x = x 0 + v 0 xt + 12 axt^2 = 0 + v 0 xt + 0 = v 0 xt
From part (a) we know that when t = 6. 2 × 10 −^2 s then x = 30 m. This allows us to solve for v 0 x:
v 0 x =
x t
30 m
= 480 m s
The muzzle velocity of the bullet is 480 m s.
(a) The motion of the beer mug is shown in Fig. 3.2(a). We choose the origin of our xy coordinate system as being at the point where the mug leaves the counter. So the mug’s initial coordinates for its flight are x 0 = 0, y 0 = 0. At the very beginning of its motion through the air, the velocity of the mug is horizontal. (This is because its velocity was horizontal all the time it was sliding on the counter.) So we know that v 0 y = 0 but we don’t know the value of v 0 x. (In fact, that’s what we’re trying to figure out!)
v o
1.40 m
0.860 m
q
q
(a) (b)
x
y
Figure 3.2: (a) Beer mug slides off counter and strikes floor! (b) Velocity vector of the beer mug at the time of impact.
We might begin by finding the time t at which the mug hit the floor. This is the time t at which y = − 0 .860 m (recall how we chose the coordinates!), and we will need the y equation of motion for this; since v 0 y = 0 and ay = −g, we get:
y = v 0 yt + 12 ayt^2 = − 12 gt^2
So we solve − 0 .860 m = − 12 gt^2
which gives
t^2 =
2(0.860 m) g
2(0.860 m) (9. 80 m s 2 )
= 0.176 s^2
so then t = 0.419 s
is the time of impact. To find v 0 x we consider the x equation of motion; x 0 = 0 and ax = 0, so we have
x = v 0 xt.
At t = 0.419 s we know that the x coordinate was equal to 1.40 m. So
1 .40 m = v 0 x(0.419 s)
Solve for v 0 x:
v 0 x =
1 .40 m 0 .419 s
= 3. 34 m s
which tells us that the initial speed of the mug was v 0 = 3. 34 m s.
(b) We want to find the components of the mug’s velocity at the time of impact, that is, at t = 0.419 s. Substitute into our expressions for vx and vy:
vx = v 0 x + axt = v 0 x = 3. 34 m s
We solve for the time at which x = 22.0 m:
x = 19. 2 m s t = 22.0 m =⇒ t =
22 .0 m
= 1.15 s
The ball hits the wall 1.15 s after being thrown.
(b) We will be able to answer this question if we can find the y coordinate of the ball at the time that it hits the wall, namely at t = 1.15 s. We need the y equation of motion. The initial y velocity of the ball is
v 0 y = v 0 sin θ 0 =
(
) sin 40. 0 ◦^ = 16. 1 m s
and the y acceleration of the ball is ay = −g giving:
y = y 0 + v 0 yt + 12 ayt^2 =
(
) t − 12 gt^2
which we use to find the y coordinate at t = 1.15 s:
y = (16. 1 m s )(1.15 s) − 12 (9. (^80) sm 2 )(1.15 s)^2 = 12.0 m
which tells us that the ball hits the wall at 12.0 m above the ground level (above the release point).
(c) The x and y components of the balls’s velocity at the time of impact, namely at t = 1.15 s are found from Eqs. 3.16:
vx = v 0 x + axt = 19. 2 m s + 0 = 19. 2 m s
and vy = v 0 y + ay t = 16. 1 m s + (− 9. 80 m s 2 )(1.15 s) = +4. 83 m s.
(d) Has the ball already passed the highest point on its trajectory? Suppose the ball was on its way downward when it struck the wall. Then the y component of the velocity would be negative, since it is always decreasing and at the trajectory’s highest point it is zero. (Of course, the x component of the velocity stays the same while the ball is in flight.) Here we see that the y component of the ball’s velocity is still positive at the time of impact. So the ball was still climbing when it hit the wall; it had not reached the highest point of its (free) trajectory.
We make a diagram of the projectile’s motion in Fig. 3.4. The launch it speed is v 0 , and the projectile is launched at an angle θ 0 upward from the horizontal. We might start this problem by solving for the time it takes the projectile to get to maxi- mum height, but we can note that at maximum height, there is no y velocity component, and
v 0
Figure 3.4: Motion of projectile in Example 8.
the x velocity component is the same as it was when the projectile was launched. Therefore at maximum height the velocity components are
vx = v 0 cos θ 0 and vy = 0
and so the speed of the projectile at maximum height is v 0 cos θ 0. Now, we are told that the launching speed (v 0 ) is five times the speed at maximum height. This gives us:
v 0 = 5v 0 cos θ 0 =⇒ cos θ 0 =
which has the solution θ 0 = 78. 5 ◦
So the elevation angle at launching is θ 0 = 78. 5 ◦.
Comment: This problem is worked in virtually every physics text, and it is sometimes simply called “The Projectile Problem”. I include it in this book for the sake of completeness and so that we can use the results if we need them later on. I do not treat it as part of the fundamental material of this chapter because it is a very particular application of free–fall motion. In this problem, the projectile impacts at the same height as the one from which it started, and that is not always the case. We must think about all projectile problems individually and not rely on simple formulae to plug numbers into! The path of the projectile is shown in Fig. 3.5. The initial coordinates of the projectile are
x 0 = 0 and y 0 = 0 ,
This is the maximum height attained by the projectile:
v^20 sin^2 θ 0 2 g
(b) What is the mathematical condition for when the projectile strikes the ground (since that is how we will find the range R)? We know that at this point, the projectile’s y coordinate is zero: y = v 0 sin θ 0 t − 12 gt^2 = 0
We want to solve this equation for t; we can factor out t in this expression to get:
t(v 0 sin θ 0 − 12 gt) = 0
which has two solutions:
t = 0 and t =
2 v 0 sin θ 0 g
The first of these is just the time when the projectile was fired; yes, y was equal to zero then, but that’s not what we want! The time at which the projectile strikes the ground is
t =
2 v 0 sin θ 0 g
We want to find the value of x at the time of impact. Substituting this value of t into our equation for x(t), we find:
x = v 0 cos θ 0
( 2 v 0 sin θ 0 g
)
2 v^20 sin θ 0 cos θ 0 g
This value of x is the range R of the projectile. We can make this result a little simpler by recalling the trig relation:
sin 2θ 0 = 2 sin θ 0 cos θ 0.
Using this in our result for the range gives:
2 v^20 sin θ 0 cos θ 0 g
v 02 sin 2θ 0 g
Now, this problem does deal with a projectile which starts and ends its flight at the same height, just as we calculated in the previous example. So we can use the results for the range R and maximum height H that we found there.
A
9.40 km B
3.30 km
35 o
Figure 3.6: Volcanic bombs away!
The problem tells us that R = 3H. Substituting the expressions for H and R that we found in the last example (we pick the first expression we got for R), we get:
2 v 02 sin θ 0 cos θ 0 g
( v^20 sin^2 θ 0 2 g
)
Cancelling stuff, we get:
2 cos θ 0 =
sin θ 0 =⇒ tan θ 0 =
The solution is:
θ 0 = tan−^1 (4/3) = 53. 1 ◦
The projectile was fired at 53. 1 ◦^ above the horizontal.
(a) We use a coordinate system with its origin at point A (the volcano “vent”); then for the flight from the vent at A to point B, the initial coordinates are x 0 = 0 and y 0 = 0, and the final coordinates are x = 9.40 km and y = − 3 .30 km. Aside from this, we don’t know the initial speed of the rock (that’s what we’re trying to find) or the time of flight from A to B. Of course, the acceleration of the rock is given by ax = 0, ay = −g. We start with the x equation of motion. The initial x−velocity is
v 0 x = v 0 cos θ
v 0
f
q 0
d
Path of projectile
Figure 3.7: Projectile is fired up an incline, as described in Example 12
(b) Having v 0 in hand, finding t is easy. Using our result from part(a) and Eq. 3.25 we find:
t =
(9.40 km) v 0 cos θ
(9400 m) (118 m s ) cos 35◦^
= 97.2 s
The time of flight is 97.2 s.
d =
2 v 02 cos θ 0 sin(θ 0 − φ) g cos^2 φ
(b) For what value of θ 0 is d a maximum, and what is the maximum value? [Ser 4-56]
(a) This is a relatively challenging problem, and of course it is completely analytic. We can start by writing down equations for x and y as functions of time. By now we can easily see that we have: x = v 0 cos θ 0 t y = v 0 sin θ 0 t − 12 gt^2
We can combine these equations to get a relation between x and y for points on the trajectory; from the first, we have t = x/(v 0 cos θ 0 ), and putting this into the second one gives:
y = v 0 sin θ 0
( x v 0 cos θ 0
) − 12 g
( x v 0 cos θ 0
) 2
= (tan θ 0 )x −
g 2
x^2 v^20 cos^2 θ 0
What is the condition for the time that the projectile hits the slope? Unlike the problems where a projectile impacts with the flat ground or a wall, we don’t know the value of x or y
at impact. But since the incline has a slope of tan φ, the relation between x and y for points on the slope is
y = (tan φ)x.
These two relations between x and y allow us to solve for the values of x and y where the impact occurs. Substituting for y above, we find:
(tan φ)x = (tan θ 0 )x −
g 2
x^2 v^20 cos^2 θ 0
A little rearranging gives:
g 2
x v^20 cos^2 θ 0
and the solution for x is:
x =
2 v^20 cos^2 θ 0 (tan θ 0 − tan φ) g
The problem has us solve for the distance d up the slope; this distance is related to the impact value of x by:
d =
x cos φ
and this gives us:
d =
x cos φ
2 v 02 cos^2 θ 0 (tan θ 0 − tan φ) g cos φ
Although this is a perfectly good expression for d, it is not the one presented in the problem. (Among other things, it has another factor of cos φ downstairs.) If we multiply top and bottom by cos φ we find:
d =
2 v^20 cos^2 θ 0 cos φ(tan θ 0 − tan φ) g cos^2 φ =
2 v^20 cos θ 0 (cos θ 0 cos φ tan θ 0 − cos θ 0 cos φ tan φ) g cos^2 φ =
2 v^20 cos θ 0 (cos φ sin θ 0 − cos θ 0 sin φ) g cos^2 φ
And now using an angle–addition identity from trigonometry in the numerator, we arrive at
d =
2 v 02 cos θ 0 sin(θ 0 − φ) g cos^2 φ
which is the preferred expression for d.
(b) In part (a) we found the up–slope impact distance as a function of launch angle θ 0. (The launch speed v 0 and the slope angle φ are taken to be fixed.) For a certain value of theta 0 ,