Preparation problems

4-5. The position of an object as a function of time is given by r = (2.4t + 1.2t2)  i + (0.89t- 1.9t2)  j m, where t is the time in seconds. What are the magnitude and direction of the acceleration?

Solution:

We begin by recalling that the acceleration is the second derivative of the position, or the first derivative of the velocity, as noted on our constants and equation sheet. So our first step is to determine the velocity of the particle:

®
v
 
 =
d ®
r
 
dt
We recall that to take the derivative of a vector, we take the derivative of each component independantly. So we get
®
v
 
=
d
dt
 é
ë 		(2.4t + 1.2t2)   ^
i
 
 + (0.89t - 1.9t2)   ^
j
 
 ù
û 		 m
=
d
dt
 (2.4t + 1.2t2)   ^
i
 
 + d
dt
 (0.89t - 1.9t2)   ^
j
 
  m
®
v
 
=
[2.4 + 2(2.4t)]   ^
i
 
 + [0.89 - 2(1.9t)]   ^
j
 
  m/s

Now we can find the acceleration recalling that

®
a
 
 =
d ®
v
 
dt
Again we do this component by component, so
®
a
 
=
d
dt
 é
ë 		(2.4 + 2.4t)   ^
i
 
 + (0.89 - 3.8t)   ^
j
 
 ù
û 		 m/s
=
d
dt
 (2.4 + 2.4t)   ^
i
 
 + d
dt
 (0.89 - 3.8t)   ^
j
 
  m/s
®
a
 
=
2.4   ^
i
 
 -3.8   ^
j
 
  m/s2

We are asked to find the magnitude and direction of the vectors, so we need recall that

a
=
  _________
Ö ax2 + ay2
 
=
  æ
Ö
(2.4  m/s2)2 + (-3.8  m/s2)2
 
a
=
4.49  m/s2
Finally, we find the direction using
q
=
tan-1 æ
ç
è 		ay
ax
 ö
÷
ø
=
tan-1 æ
ç
è 		-3.8  m/s2
2.4  m/s2
 ö
÷
ø
q
=
-57.7°

4-18. You are trying to roll a ball off a 80.0-cm-high table to squash a bug on the floor 50.0 cm from the table's edge. How fast should you roll the ball?

Solution:

We clearly have a constant acceleration two-dimensional motion problem. The ball will be subject to a constant acceleration from gravity in the vertical direction, and is initially moving only in the horizontal direction with no horizontal acceleration. We want to find the initial velocity of the ball, or v0 = vx0 [^(i)] + vy0 [^(j)]. Specifically, we are looking for the x-component of the initial velocity, or vx0. As always, we should write down what we know:

Starting Conditions Value
Time t0 0 s
Positions x0 (Edge of table) 0 m
y0 0.8 m
Velocities vx0 unknown
vy0 0 m/s
Accelerations ax 0 m/s2
ay -g = -9.8 m/s2
Ending Conditions
Position x 0.5 m
y 0 m

We can find the initial velocity of the ball using the equation

x = x0 + vx0 t + 1
2
 ax t2 .
We can simplify by substituting in zero values and rearrange it to get an equation for vx0
vx0 = x
t
 .
(1)
Again, we do not know the time the ball spends in the air, so we must turn to the other component of the motion to find this information. The motion in the y-direction is given by the equation
y = y0 + vy0 t + 1
2
 ay t2 .
Since y = 0, vy0 = 0, and ay = -g, we can simplify this to
t =   æ
Ö
2 y0
g
 
 .
(2)
Substituting Eq. (2) into Eq. (1), we get a nasty looking equation for vx0, but we can plug in our known values to get an answer:
vx0
=
x
  æ
Ö
2 y0
g
 
=
0.5  m
  æ
Ö
2 ×0.8  m
9.8  m/s2
 
vx0
=
1.24  m/s

4-28. Compare the travel times for the projectiles launched at 30° and 60° in Fig. 4-13 (on page 75), both of which have the same starting and ending points.

Solution:

Looking at Fig. 4-13, we see that the caption tells us that the initial speed of the projectile for each case is v0 = 50 m/s. Also, we don't know the exact final position of the projectile along the x-axis, just that it is the same in each case. We therefore have two unknowns: the time, t; and the ending horizontal position, x. Common sense tells us that the higher angle should take longer to hit the ground, so we will see if that holds. Let's start by writing down a table of what we know, which might look something like this:

Final Time t =  ?
Speed v0 = 50 m/s
x-direction y-direction
Initial Positions x0 = 0 m y0 = 0 m
Velocities vx0 = v0 cosq vy0 = v0 sinq
Accelerations ax = 0 m/s2 ay = -g = -9.8 m/s2
Final Positions x =  ? y = 0 m

Now we recall that when we solve 2-D kinematic equations, we typically do not want to use the third of the kinematic equations from our constants and equation sheet. Looking at the other two, we see that using the first equation in the y-direction will leave us with only one unknown, t. The equation looks like

y = y0 + vy0 t + 1
2
 ay t2 .
Substituting in known values except for the angle, q, we can get a simplified form of this equation, and then solve for time in terms of the q.
(0  m)
=
(0  m) + v0 sinqt - 1
2
 g t2
1
2
 g t2
=
v0 sinqt
g t2
=
2 v0 sinqt
t
=
2 v0 sinq
g
Using this equation, we can substitute in and solve for each angle.

For q = 30°:

t
=
2 v0 sinq
g
t
=
2 (50  m/s) sin(30°)
9.8  m/s2
t
= 5.10  s

For q = 60°:

t
=
2 v0 sinq
g
t
=
2 (50  m/s) sin(60°)
9.8  m/s2
t
= 8.83  s

As we expected, the higher angle took longer to hit the ground.

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Solutions translated from TEX by TTH, version 1.57.