Preparation problems

2-51. Amtrak's 20th-Century Limited is en route from Chicago to New York at 110 km/h, when the engineer spots a cow on the track. The train brakes to a halt in 1.2 min with a constant deceleration, stopping just in front of the cow. (a) What is the magnitude of the train's acceleration? (b) What is the direction of the acceleration? (c) How far was the train from the cow when the engineer first applied the brakes?

Solution:

We know the following information to start out:

Initial Position
x0
= 0  m
Initial Velocity:
vx0
= 110  km/h = 30.56  m/s
Acceleration:
ax
= ?
Final Velocity
vx
= 0  km/h = 0  m/s
Final Position
x
= ?
Final Time
t
= 1.2  min = 72  s

(a) We need to find an equation that will give us the acceleration knowing the initial and final velocities, and the time of acceleration. Looking at our constants and equation sheet, we see that

vx = vx0 + ax t
should get us the answer. Solving this equation for ax and plugging in our known values, we get the answer:
ax
=
vx - vx0
t
=
0  m/s - 30.56  m/s
72  s
ax
=
-0.42  m/s2 .
The question asks for the magnitude, so the final answer is ax = 0.42  m/s2.

(b) Clearly, since the train slowed down (the acceleration in part (a) was negative), the direction of the acceleration is opposite the direction of the train, or back towards Chicago.

(c) We need to use another of our kinematics equations to solve this portion of the problem. Either of the other kinematics equations on the constants and equation sheet will allow us to get the answer. I will use the first equation:

x = x0 + vx0 t + 1
2
 ax t2 .
Our only unkown here is x, so plugging in the other values, we get our answer,
x
=
0  m + (30.56  m/s)(72  s) + 1
2
 (-0.42  m/s2) (72  s)2
=
1100  m

2-52. A jetliner touches down at 220 km/h, reverses its engines to provide braking, and comes to a halt 29 s later. What is the shortest runway on which this aircraft can land, assuming constant deceleration starting at touchdown?

Solution:

We know the following information to start out:

Initial Position
x0
= 0  m
Initial Velocity:
vx0
= 220  km/h = 61.12  m/s
Acceleration:
ax
= ?
Final Velocity
vx
= 0  km/h = 0  m/s
Final Position
x
= ?
Final Time
t
= 29  s

We need to find a way to determine the final position of the plane. We clearly do not have enough information to use the first or third equation on the constants and equation sheet directly. Therefore, we must first find another unknown using the second equation,

vx = vx0 + ax t .
It appears we can use this equation to find a value for the acceleration. Once we have this, we can use either of the other equations to find our final position. Solving for the acceleration, we get
ax
=
vx - vx0
t
=
0  m/s - 61.12  m/s
29  s
ax
=
-2.11  m/s2 .

We now need to use another of our kinematics equations to solve this portion of the problem. Either of the other kinematics equations on the constants and equation sheet will allow us to get the answer. I will use the first equation:

x = x0 + vx0 t + 1
2
 ax t2 .
Our only unkown here is x, so plugging in the other values, we get our answer,
x
=
0  m + (61.12  m/s)(29  s) + 1
2
 (-2.11  m/s2) (29  s)2
=
886.1  m



Follow-up problems

2-20. For the motion plotted in Fig. 2-19 (on page 41), estimate (a) the greatest velocity in the positive x direction; (b) the greatest velocity in the negative x direction; (c) any times when the object is instantaneously at rest; and (d) the average velocity over the interval shown.

Solution:

Since we do not know the exact function that describes the motion pictured, we must estimate the velocity. We can do this recalling that the instantaneous velocity is the slope of the position vs. time curve.

(a) We see that the maximum positive slope occurs at about 2 s, and is approximately 3 m/s.

(b) We see that the maximum negative slope occurs at about 3.6 s, and is approximately -1.7 m/s.

(c) We see that the slope of the curve is zero at two places, and therefore the object is instantaneously at rest at t=3 s and t=5 s.

(d) The average velocity over the interval shown is found by taking

vav(0 ® 6  s)
=
Dx
Dt
=
x(6  s) - x0
6  s - 0  s
=
3  m - 0  m
6  s
vav(0 ® 6  s)
=
0.5  m/s

2-33. Your plane reaches its takeoff runway and then holds for 4.0 min because of air-traffic congestion. The plane then heads down the runway with an average acceleration of 3.6 m/s2. It is airborne 35 s later. What are (a) its takeoff speed and (b) its average acceleration from the time it reaches the takeoff runway until it's airborne?

Solution:

(a) We can use the kinematics equation from our constants and equation sheet

vx = vx0 + ax t
(1)
to solve this problem. For part (a), we can say that our zero time occurs when the plane starts accelerating. We then know that t = 35 s is our final time, our acceleration over this period is ax = 3.6 m/s2, and our initial velocity is vx0 = 0 m/s. Our equation then looks like:
vx = (0  m/s) + (3.6  m/s2)(35  s) .
This gives us a takeoff speed of vx = 126 m/s.

(b) To find the average acceleration over the entire time, we need to use the equation:

aav = Dv
Dt
 .
(2)
We know our initial time is -4 min = -240 s. Our final time is 35 s. The velocity at t = -240 s is v(-240  s) = 0 m/s, while the velocity at t = 35 s is v(35  s) = 126 m/s. Our equation then looks like:
aav = v(35  s) - v(-240  s )
35  s - (-240  s)
 = 126  m/s
275  s
 = 0.46  m/s2

2-36. The position of an object is given by x = bt3, where x is in meters, t in seconds, and where the constant b is 1.5 m/s3. Determine (a) the instantaneous velocity and (b) the instantaneous acceleration at the end of 2.5 s. Find (c) the average velocity and (d) the average acceleration during the first 2.5 s.

Solution:

(a) We need to recall the following definition from our constants and equation sheet:

v(t) = d x
d t
 .
We see immediately that we can find the velocity at 2.5 s by taking the derivative of x, or
v(t)
=
d
dt
 [bt3]
=
3bt2
v(2.5  s)
=
3 (1.5  m/s3) (2.5  s)2
=
28.125  m/s

(b) We next can recall that the instantaneous acceleration is given by the following definition from our constants and equation sheet:

a(t) = d v
d t
 .
We see immediately that we must take another derivative, giving us
a(t)
=
d
dt
 [3bt2]
=
6bt
a(2.5  s)
=
6 (1.5  m/s3) (2.5  s)
=
22.5  m/s2

(c) To find the average velocity during the first 2.5 s, we recall:

vav
=
Dx
Dt
=
x(2.5  s) - x(0  s)
2.5  s
We can find x(2.5 s) using our given formula x = bt3. Using this, we get
vav
=
[(1.5  m/s3) (2.5  s)3] - 0  m
2.5  s
=
9.375  m/s

(d) To find the average acceleration during the first 2.5 s, we recall:

aav
=
Dv
Dt
=
v(2.5  s) - v(0  s)
2.5  s
We can find v(2.5 s) using the formula we got in part (b), v = 3bt2. Using this, we get
aav
=
[3 (1.5  m/s3) (2.5  s)2] - 0  m/s
2.5  s
=
11.25  m/s2

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