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Chapter 2
Motion in One Dimension
2.1 The Important Stuff
2.1.1 Position, Time and Displacement
We begin our study of motion by considering objects which are very small in comparison to the size of their movement through space. When we can deal with an object in this way we refer to it as a particle. In this chapter we deal with the case where a particle moves along a straight line. The particle’s location is specified by its coordinate, which will be denoted by x or y. As the particle moves, its coordinate changes with the time, t. The change in position from x 1 to x 2 of the particle is the displacement ∆x, with ∆x = x 2 − x 1.
2.1.2 Average Velocity and Average Speed
When a particle has a displacement ∆x in a change of time ∆t, its average velocity for that time interval is
v =
∆x ∆t
x 2 − x 1 t 2 − t 1
The average speed of the particle is absolute value of the average velocity and is given by
s =
Distance travelled ∆t
In general, the value of the average velocity for a moving particle depends on the initial and final times for which we have found the displacements.
2.1.3 Instantaneous Velocity and Speed
We can answer the question “how fast is a particle moving at a particular time t?” by finding the instantaneous velocity. This is the limiting case of the average velocity when the time
28 CHAPTER 2. MOTION IN ONE DIMENSION
interval ∆t include the time t and is as small as we can imagine:
v = lim ∆t→ 0
∆x ∆t
dx dt
The instantaneous speed is the absolute value (magnitude) of the instantaneous ve- locity.
If we make a plot of x vs. t for a moving particle the instantaneous velocity is the slope of the tangent to the curve at any point.
2.1.4 Acceleration
When a particle’s velocity changes, then we way that the particle undergoes an acceleration. If a particle’s velocity changes from v 1 to v 2 during the time interval t 1 to t 2 then we define the average acceleration as
v =
∆x ∆t
x 2 − x 1 t 2 − t 1
As with velocity it is usually more important to think about the instantaneous accel- eration, given by
a = lim ∆t→ 0
∆v ∆t
dv dt
If the acceleration a is positive it means that the velocity is instantaneously increasing; if a is negative, then v is instantaneously decreasing. Oftentimes we will encounter the word deceleration in a problem. This word is used when the sense of the acceleration is opposite that of the instantaneous velocity (the motion). Then the magnitude of acceleration is given, with its direction being understood.
2.1.5 Constant Acceleration
A very useful special case of accelerated motion is the one where the acceleration a is constant. For this case, one can show that the following are true:
v = v 0 + at (2.6) x = x 0 + v 0 t + 12 at^2 (2.7) v^2 = v 02 + 2a(x − x 0 ) (2.8) x = x 0 + 12 (v 0 + v)t (2.9)
In these equations, we mean that the particle has position x 0 and velocity v 0 at time t = 0; it has position x and velocity v at time t.
These equations are valid only for the case of constant acceleration.
30 CHAPTER 2. MOTION IN ONE DIMENSION
(a) This is not straight line motion of course, but we can sill find an average speed by dividing the distance traveled (around a circular path) by the time interval. Here, the distance traveled by the Earth as it goes once around the Sun is the circumference of the orbit, C = 2πR = 2π(1. 5 × 108 km) = 9. 42 × 108 km = 9. 42 × 1011 m
and the time interval over which that takes place is one year,
1 yr = 365.25 day
( 24 hr 1 day
) ( 3600 s 1 hr
) = 3. 16 × 107 s
so the average speed is
s =
C
t
- 42 × 1011 m
- 16 × 107 s
= 2. 99 × 10 4 m s
(b) To convert this to mi s , use 1 mi = 1.609 km. Then
s =
(
- 99 × 10 4 m s
) (^ 1 mi
- 609 × 103 m
) = 18. 6 mi s
2.2.2 Acceleration
- An electron moving along the x axis has a position given by x = (16te−t) m, where t is in seconds. How far is the electron from the origin when it momentarily stops? [HRW6 2-20]
To find the velocity of the electron as a function of time, take the first derivative of x(t):
v =
dx dt
= 16e−t^ − 16 te−t^ = 16e−t(1 − t) m s
again where t is in seconds, so that the units for v are m s. Now the electron “momentarily stops” when the velocity v is zero. From our expression for v we see that this occurs at t = 1 s. At this particular time we can find the value of x:
x(1 s) = 16(1)e−^1 m = 5.89 m
The electron was 5.89 m from the origin when the velocity was zero.
- (a) If the position of a particle is given by x = 20t − 5 t^3 , where x is in meters and t is in seconds, when if ever is the particle’s velocity zero? (b) When is its acceleration a zero? (c) When is a negative? Positive? (d) Graph x(t), v(t), and a(t). [HRW5 2-28]
2.2. WORKED EXAMPLES 31
(a) From Eq. 2.3 we find v(t) from x(t):
v(t) =
dx dt
d dt
(20t − 5 t^3 ) = 20 − 15 t^2
where, if t is in seconds then v will be in m s. The velocity v will be zero when
20 − 15 t^2 = 0
which we can solve for t:
15 t^2 = 20 =⇒ t^2 =
= 1.33 s^2
(The units s^2 were inserted since we know t^2 must have these units.) This gives:
t = ± 1 .15 s
(We should be careful... t may be meaningful for negative values!)
(b) From Eq. 2.5 we find a(t) from v(t):
a(t) =
dv dt
d dt
(20 − 15 t^2 ) = − 30 t
where we mean that if t is given in seconds, a is given in m s 2. From this, we see that a can be zero only at t = 0.
(c) From the result is part (b) we can also see that a is negative whenever t is positive. a is positive whenever t is negative (again, assuming that t < 0 has meaning for the motion of this particle).
(d) Plots of x(t), v(t), and a(t) are given in Fig. 2.1.
- In an arcade video game a spot is programmed to move across the screen according to x = 9. 00 t − 0. 750 t^3 , where x is distance in centimeters measured from the left edge of the screen and t is time in seconds. When the spot reaches a screen edge, at either x = 0 or x = 15.0 cm, t is reset to 0 and the spot starts moving again according to x(t). (a) At what time after starting is the spot instantaneously at rest? (b) Where does this occur? (c) What is its acceleration when this occurs? (d) In what direction is it moving just prior to coming to rest? (e) Just after? (f) When does it first reach an edge of the screen after t = 0? [HRW5 2-31]
(a) This is a question about the instantaneous velocity of the spot. To find v(t) we calculate:
v(t) =
dx dt
d dt
(9. 00 t − 0. 750 t^3 ) = 9. 00 − 2. 25 t^2
where this expression will give the value of v in cm s when t is given in seconds.
2.2. WORKED EXAMPLES 33
We want to know the value of t for which v is zero, i.e. the spot is instantaneously at rest. We solve:
- 00 − 2. 25 t^2 = 0 =⇒ t^2 =
= 4.00 s^2
(Here we have filled in the proper units for t^2 since by laziness they were omitted from the first equations!) The solutions to this equation are
t = ± 2 .00 s
but since we are only interested in times after the clock starts at t = 0, we choose t = 2.00 s.
(b) In this part we are to find the value of x at which the instantaneous velocity is zero. In part (a) we found that this occurred at t = 3.00 s so we calculate the value of x at t = 2.00 s:
x(2.00 s) = 9. 00 · (2.00) − 0. 750 · (2.00)^3 = 12.0 cm
(where we have filled in the units for x since centimeters are implied by the equation). The dot is located at x = 12.0 cm at this time. (And recall that the width of the screen is 15 .0 cm.)
(c) To find the (instantaneous) acceleration at all times, we calculate:
a(t) =
dv dt
d dt
(9. 00 − 2. 25 t^2 ) = − 4. 50 t
where we mean that if t is given in seconds, a will be given in m s 2. At the time in question (t = 2.00 s) the acceleration is
a(t = 2.00 s) = − 4. 50 · (2.00) = − 9. 00
that is, the acceleration at this time is − 9. 00 m s 2.
(d) From part (c) we note that at the time that the velocity was instantaneously zero the acceleration was negative. This means that the velocity was decreasing at the time. If the velocity was decreasing yet instantaneously equal to zero then it had to be going from positive to negative values at t = 2.00 s. So just before this time its velocity was positive.
(e) Likewise, from our answer to part (d) just after t = 2.00 s the velocity of particle had to be negative.
(f) We have seen that the dot never gets to the right edge of the screen at x = 15.0 cm. It will not reverse its velocity again since t = 2.00 s is the only positive time at which v = 0. So it will keep moving to back to the left, and the coordinate x will equal zero when we have:
x = 0 = 9. 00 t − 0. 750 t^3
Factor out t to solve:
t(9. 00 − 0. 750 t^2 ) = 0 =⇒
{ (^) t = 0 or (9. 00 − 0. 750 t^2 ) = 0 otherwise.
34 CHAPTER 2. MOTION IN ONE DIMENSION
-10.
-5.
0.
5.
10.
x, cm 15.
-1 0 1 2 3 4 t, s
Figure 2.2: Plot of x vs t for moving spot. Ignore the parts where x is negative!
The first solution is the time that the dot started moving, so that is not the one we want. The second case gives:
(9. 00 − 0. 750 t^2 ) = 0 =⇒ t^2 =
= 12.0 s^2
which gives t = 3.46 s
since we only want the positive solution. So the dot returns to x = 0 (the left side of the screen) at t = 3.46 s. If we plot the original function x(t) we get the curve given in Fig. 2.2 which shows that the spot does not get to x = 15.0 cm before it turns around. (However as explained in the problem, the curve does not extend to negative values as the graph indicates.)
2.2.3 Constant Acceleration
- The head of a rattlesnake can accelerate 50 m s 2 in striking a victim. If a car could do as well, how long would it take to reach a speed of 100 km hr from rest? [HRW5 2-33]
First, convert the car’s final speed to SI units to make it easier to work with:
km hr
( 100
km hr
) ·
( (^) 1000 m
1 km
) ·
( 1 hr 3600 s
) = 27. 8 m s
The acceleration of the car is 50 m s 2 and it starts from rest which means that v 0 = 0. As we’ve found, the final velocity v of the car is 27. 8 m s. (The problem actually that this is final
36 CHAPTER 2. MOTION IN ONE DIMENSION
100 m/s (^) v = 0
x
a = -5.0 m/s^2
Figure 2.3: Plane touches down on runway at 100 m s and comes to a halt.
The plane needs 20 s to come to a halt.
(b) The plane also travels the shortest distance in stopping if its acceleration is − 5. 0 m s 2. With x 0 = 0, we can find the plane’s final x coordinate using Eq. 2.9, using t = 20 s which we got from part (a):
x = x 0 + 12 (v 0 + v)t = 0 + 12 (100 m s + 0)(20 s) = 1000 m = 1.0 km
The plane must have at least 1.0 km of runway in order to come to a halt safely. 0.80 km is not sufficient.
- A drag racer starts her car from rest and accelerates at 10. 0 m s 2 for the entire distance of 400 m ( 14 mile). (a) How long did it take the car to travel this distance? (b) What is the speed at the end of the run? [Ser4 2-33]
(a) The racer moves in one dimension (along the x axis, say) with constant acceleration a = 10. 0 m s 2. We can take her initial coordinate to be x 0 = 0; she starts from rest, so that v 0 = 0. Then the location of the car (x) is given by:
x = x 0 + v 0 t + 12 at^2 = = 0 + 0 + 12 at^2 = 12 (10. 0 m s 2 )t^2
We want to know the time at which x = 400 m. Substitute and solve for t:
400 m = 12 (10. 0 m s 2 )t^2 =⇒ t^2 =
2(400 m) (10. 0 m s 2 )
= 80.0 s^2
which gives t = 8.94 s.
The car takes 8.94 s to travel this distance.
(b) We would like to find the velocity at the end of the run, namely at t = 8.94 s (the time we found in part (a)). The velocity is:
v = v 0 + at = 0 + (10. (^0) sm 2 )t = (10. 0 m s 2 )t
2.2. WORKED EXAMPLES 37
1.0 cm
Path of electron
Voltage source
Accelerating region
Figure 2.4: Electron is accelerated in a region between two plates, in Example 10.
At t = 8.94 s, the velocity is
v = (10. 0 m s 2 )(8.94 s) = 89. 4 m s
The speed at the end of the run is 89. 4 m s.
- An electron with initial velocity v 0 = 1. 50 × 10 5 m s enters a region 1 .0 cm long where it is electrically accelerated, as shown in Fig. 2.4. It emerges with velocity v = 5. 70 × 10 6 m s. What was its acceleration, assumed constant? (Such a process occurs in the electron gun in a cathode–ray tube, used in television receivers and oscilloscopes.) [HRW5 2-39]
We are told that the acceleration of the electron is constant, so that Eqs. 2.6–2.9 can be used. Here we know the initial and final velocities of the electron (v 0 and v). If we let its initial coordinate be x 0 = 0 then the final coordinate is x = 1.0 cm = 1. 0 × 10 −^2 m. We don’t know the time t for its travel through the accelerating region and of course we don’t know the (constant) acceleration, which is what we’re being asked in this problem. We see that we can solve for a if we use Eq. 2.8:
v^2 = v 02 + 2a(x − x 0 ) =⇒ a =
v^2 − v 02 2(x − x 0 )
Substitute and get:
a =
(5. 70 × 10 6 m s )^2 − (1. 50 × 10 5 m s )^2 2(1. 0 × 10 −^2 m) = 1. 62 × 10 15 m s 2
The acceleration of the electron is 1. 62 × 10 15 m s 2 (while it is in the accelerating region).
2.2. WORKED EXAMPLES 39
22.2 m/s
a = -2.1 m/s^2 6.94 m/s
1 2
50 m
x 2
x 1
Figure 2.5: Two subway trains in Example 12.
The trains will collide if there is ever a time at which their coordinates are equal. So we want to see if there is a t which gives the condition:
(22. 2 m s )t + (− 1. 05 m s 2 )t^2 = 50 m + (6. 94 m s )t
This is a quadratic equation, for which we can use the quadratic formula. Neglecting the units for simplicity, we can rearrange the terms and rewrite it as
- 05 t^2 − 15. 28 t + 50 = 0
and the quadratic formula gives the answers as
−b ±
b^2 − 4 ac 2 a
√ (15.28)^2 − 4(1.05)(50) 2(1.05)
{ 9 .58 s 4 .97 s
This is a little confusing because there are two possible answers! (Both values of t are positive.) But the answer we want is the first one, 4.97 s — after the collision, the second time is not relevant^1. So the trains will collide t = 4.97 s after the rear car begins to decelerate. At the time we have found, the velocity of the rear train is
v = v 0 + at = 22. 2 m s + (− 2. 1 m s 2 )(4.97 s) = 11. 8 m s
and the velocity of the front train remains 6. 94 m s. So at the time of the collision, the rear train is going faster by a difference of
∆v = 11. 8 m s − 6. 94 m s = 4. 8 m s
That is the relative speed at which the collision takes place.
2.2.4 Free Fall
40 CHAPTER 2. MOTION IN ONE DIMENSION
y
v 0
y = 0 m
y =50 m
Figure 2.6: Object thrown upward reaches height of 50 m.
- (a) With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 50 m? (b) How long will it be in the air? [HRW5 2-61]
(a) First, we decide on a coordinate system. I will use the one shown in Fig. 2.6, where the y axis points upward and the origin is at ground level. The ball starts its flight from ground level so its initial position is y 0 = 0. When the ball is at maximum height its coordinate is y = 50 m, but we also know its velocity at this point. At maximum height the instantaneous velocity of the ball is zero. So if our “final” point is the time of maximum height, then v = 0. So for the trip from ground level to maximum height, we know y 0 , y, v and the acceleration a = − 9. 8 m s 2 = −g, but we don’t know v 0 or the time t to get to maximum height. From our list of constant–acceleration equations, we see that Equation 2.8 will give us the initial velocity v 0 :
v^2 = v^20 + 2a(y − y 0 ) =⇒ v^20 = v^2 − 2 a(y − y 0 )
Substitute, and get: v^20 = (0)^2 − 2(− 9. 8 m s 2 )(50 m − 0) = 980 m 2 s^2
The next step is to “take the square root”. Since we know that v 0 must be a positive number, we know that we should take the positive square root of 980 m
2 s^2. We get:
v 0 = +31 m s
The initial speed of the ball is 31 m s
(b) We want to find the total time that the ball is in flight. What do we know about the ball when it returns to earth and hits the ground? We know that its y coordinate is equal to zero. (So far, we don’t know anything about the ball’s velocity at the the time it returns to ground level.) If we consider the time between throwing and impact, then we do know y 0 , y, v 0 and of course a. If we substitute into Eq. 2.7 we find:
(^1) However it would be relevant if the trains were on parallel tracks; then the collision would not take place
and we could find the times at which they were side-by-side and their relative velocities at those times.
42 CHAPTER 2. MOTION IN ONE DIMENSION
y^ v^0
4.00 m
Figure 2.8: Student throws her keys into the air, in Example 15.
But at the time of impact we have
y = − 30 .0 m = (− 8. 00 m s )t − 12 gt^2 = (− 8. 00 m s )t − (4. (^90) sm 2 )t^2 ,
an equation for which we can solve for t. We rewrite it as:
(4. 90 m s 2 )t^2 + (8. 00 m s )t − 30 .0 m = 0
which is just a quadratic equation in t. From our algebra courses we know how to solve this; the solutions are:
t =
−(8. 00 m s ) ±
√ (8. 00 m s )^2 − 4(4. 90 m s 2 )(− 30 .0 m) 2(4. 90 m s 2 )
and a little calculator work finally gives us:
t =
{ − 3 .42 s 1 .78 s
Our answer is one of these... which one? Obviously the ball had to strike the ground at some positive value of t, so the answer is t = 1.78 s. The ball strikes the ground 1.78 s after being thrown.
- A student throws a set of keys vertically upward to her sorority sister in a window 4 .00 m above. The keys are caught 1 .50 s later by the sister’s outstretched hand. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught? [Ser4 2-47]
(a) We draw a simple picture of the problem; such a simple picture is given in Fig. 2.8. Having a picture is important, but we should be careful not to put too much into the picture; the problem did not say that the keys were caught while they were going up or going down. For all we know at the moment, it could be either one!
2.2. WORKED EXAMPLES 43
We will put the origin of the y axis at the point where the keys were thrown. This simplifies things in that the initial y coordinate of the keys is y 0 = 0. Of course, since this is a problem about free–fall, we know the acceleration: a = −g = − 9. 80 m s 2. What mathematical information does the problem give us? We are told that when t = 1.50 s, the y coordinate of the keys is y = 4.00 m. Is this enough information to solve the problem? We write the equation for y(t):
y = y 0 + v 0 t + 12 at^2 = v 0 t − 12 gt^2
where v 0 is presently unknown. At t = 1.50 s, y = 4.00 m, so:
4 .00 m = v 0 (1.50 s) − 12 (9. 80 m s 2 )(1.50 s)^2.
Now we can solve for v 0. Rearrange this equation to get:
v 0 (1.50 s) = 4.00 m + 12 (9. (^80) sm 2 )(1.50 s)^2 = 15.0 m.
So:
v 0 =
15 .0 m 1 .50 s
= 10. 0 m s
(b) We want to find the velocity of the keys at the time they were caught, that is, at t = 1.50 s. We know v 0 ; the velocity of the keys at all times follows from Eq. 2.6,
v = v 0 + at = 10. 0 m s − 9. (^80) sm 2 t
So at t = 1.50 s, v = 10. 0 m s − 9. (^80) sm 2 (1.50 s) = − 4. 68 m s.
So the velocity of the keys when they were caught was − 4. 68 m s. Note that the keys had
a negative velocity; this tells us that the keys were moving downward at the time they were caught!
- A ball is thrown vertically upward from the ground with an initial speed of
- 0 m s. (a) How long does it take the ball to reach its maximum altitude? (b) What is its maximum altitude? (c) Determine the velocity and acceleration of the ball at t = 2.00 s. [Ser4 2-49]
(a) An illustration of the data given in this problem is given in Fig. 2.9. We measure the coordinate y upward from the place where the ball is thrown so that y 0 = 0. The ball’s acceleration while in flight is a = −g = − 9. 80 m s 2. We are given that v 0 = +15. 0 m s. The ball is at maximum altitude when its (instantaneous) velocity v is zero (it is neither going up nor going down) and we can use the expression for v to solve for t:
v = v 0 + at =⇒ t =
v − v 0 a
2.2. WORKED EXAMPLES 45
v 0
t = 3.00 s v=
Figure 2.10: Ball is hit straight up; reaches maximum height 3.00 s later.
(a) An illustration of the data given in the problem is given in Fig. 2.10. For the period from when the ball is hit to the time it reaches maximum height, we know the time interval, the acceleration (a = −g) and also the final velocity, since at maximum height the velocity of the ball is zero. Then Eq. 2.6 gives us v 0 :
v = v 0 + at =⇒ v 0 = v − at
and we get: v 0 = 0 − (− 9. (^80) sm 2 )(3.00 s) = 29. 4 m s
The initial velocity of the ball was +29. 4 m s.
(b) To find the value of the maximum height, we need to find the value of the y coordinate at time t = 3.00 s. We can use either Eq. 2.7 or Eq. 2.8. the latter gives:
v^2 = v^20 + 2a(y − y 0 ) =⇒ (y − y 0 ) =
v^2 − v 02 2 a
Plugging in the numbers we find that the change in y coordinate for the trip up was:
y − y 0 =
02 − (29. 4 m s )^2 2(− 9. 80 m s 2 )
= 44.1 m.
The ball reached a maximum height of 44 .1 m.
- A parachutist bails out and freely falls 50 m. Then the parachute opens, and thereafter she decelerates at 2. 0 m s 2. She reaches the ground with a speed of 3. 0 m s. (a) How long was the parachutist in the air? (b) At what height did the fall begin? [HRW5 2-84]
(a) This problem gives several odd bits of information about the motion of the parachutist! We organize the information by drawing a diagram, like the one given in Fig. 2.11. It is
46 CHAPTER 2. MOTION IN ONE DIMENSION
(a)
(b)
(c)
v=
v=3.0 m/s
Free Fall 50 m
Deceleration
Figure 2.11: Diagram showing motion of parachutist in Example 18.
very important to organize our work in this way! At the height indicated by (a) in the figure, the skydiver has zero initial speed. As she falls from (a) to (b) her acceleration is that of gravity, namely 9. 80 m s 2 downward. We know that (b) is 50 m lower than (a) but we don’t yet know the skydiver’s speed at (b). Finally, at point (c) her speed is 3. 0 m s and between (b) and (c) her “deceleration” was 2. 0 m s 2 , but we don’t know the difference in height between (b) and (c). How can we start to fill in the gaps in our knowledge? We note that on the trip from (a) to (b) we do know the starting velocity, the distance travelled and the acceleration. From Eq. 2.8 we can see that this is enough to find the final velocity, that is, the velocity at (b). Use a coordinate system (y) which has its origin at level (b), and the y axis pointing upward. Then the initial y coordinate is y 0 = 50 m and the the initial velocity is v 0 = 0. The final y coordinate is y = 0 and the acceleration is constant at a = − 9. 80 m s 2. Then using Eq. 2.8 we have:
v^2 = v^20 + 2a(y − y + 0) = 0 + 2(− 9. 80 m s 2 )(0 − 50 m) = 980 m
2 s^2
which has the solutions v = ± 31. 3 m s
but here the skydiver is obviously moving downward at (b), so we must pick
v = − 31. 3 m s
for the velocity at (b). While we’re at it, we can find the time it took to get from (a) to (b) using Eq. 2.6, since we know the velocities and the acceleration for the motion. We find:
v = v 0 + at =⇒ t =
v − v 0 a
