Except for a little graphing, we have dealt only with constant velocity. Although this is simple mathematically, it is physically very limiting and so we will now extend our scope to encompass acceleration. We begin by defining it:

		change in velocity		delta v
acceleration	=	-----------------	=	------
		change in time		delta t

Our study will cover only uniform acceleration that is, only those cases in which the velocity changes at a steady rate. Notice the units of acceleration:

		delta v		(length/time)
a (acceleration)	=	-----------	or	---------------
		delta t		time

This can be expressed in SI units as m/sec/sec or m/sec², and read as ";meters per second per second"; or ";meters per second squared."; In English units, ft/sec/sec or ft/sec². In general, length/time/time or length/time². Think of 7 cm/sec² as being a gain of 7 cm/sec per second. That is, in one second, a gain of 7 cm/sec, in two seconds a gain of 14 cm/sec, etc.

As before, we will begin without regard for the direction (or vector nature) of acceleration, though this will become quite important later.

50. A car starts from rest and reaches a speed of 8.4 m/sec in 5.9 seconds. What is its acceleration?

a = Çv / Çt a = (8.4 m/sec - 0 m/sec) / 5.9 sec a = 1.42 m/sec2 That is, the velocity increases by 1.42 m/sec each second.

51. A car traveling at 12 ft/sec accelerates at 2 ft/sec². How fast is it traveling in
a) 1 second?
b) 2 seconds?
c) 5 seconds?
d) t seconds?

a) a = v/t v = at v = at + v0 v = (2 ft/sec2)(1 sec) + 12 ft/sec v = 14 ft/sec

b) v = at + v0 v = (2 ft/sec2)(2 sec) + 12 ft/sec v = 16 ft/sec

c) v = at + v0 v = (2 ft/sec2)(5 sec) + 12 ft/sec v = 22 ft/sec

d) v = at + v0 v = (2 ft/sec2)(t sec) + 12 ft/sec v = 12 ft/sec + 2t

52. Rusty Nails slows his bike from 6.6 m/sec to 1.4 m/sec in 2.6 seconds. What is his acceleration?

A negative acceleration: a = Çv / Çt a = (1.4 - 6.6) m/sec / 2.6 sec a = -2.0 m/sec2 In this case indicates the bicycle is slowing down.

NOTE: To find how much a value has changed, subtract the original value from the final value.

53. Milton Yermouth invents a toy car that accelerates uniformly at 7.2 cm/sec². Starting from rest, how fast will it be going in 3.0 seconds?

a = Çv / Çt 2) (3.0 sec) Çv = 21.6 cm/sec [v = at for easy memorization]

54. The bolt of a crossbow accelerates at 224 m/sec² and attains a speed of 131 m/sec when it leaves the bow. For how long is it accelerated?

a = Çv / Çt Çt = Çv / a Çt = (131 m/sec) / (224 m/sec2) Çt = .58 sec

55. A plane traveling at 215 m/sec accelerates at 23 m/sec² for 9.3 seconds. What is the final velocity?

a = Çv / Çt Çv = aÇt (or, as previously mentioned, v = at) But wait! we have an intial velocity v0. Any additional velocity is over and above this initial value. Hence: v = at + v0 v = (23 m/sec2) (9.3 sec) + 215 m/sec v = 429 m/sec

In two-dimensional motion, each component of motion acts independently of the other. Because of this we may treat motions in the x and y directions in separate equations. Indeed, we must treat them separately, or we will not reflect the reality of nature.

56*. A particle travels to the right at a constant rate of 7.0 m/sec. It moves vertically up at 2.6 m/sec², starting from rest.
a) What is its direction of travel after 4.4 seconds?
b) What is its speed at this time?

a) horizontal velocity: vx = 7.0 m/sec velocity velocity: vy = ayt vy = (2.6 m/sec2) (4.4 sec) vy = 11.4 m/sec tanØ = 11.4 / 7.0 Ø = 58.5°

b) R = (11.72 - 7.02)1/2 R = 13.4 m/sec

57*. A particle travels to the right with an acceleration of 2.6 m/sec² and up at 3.4 m/sec². If it started from rest, what are its speed and direction after 8.7 seconds?

Notice that the direction of motion depends only on the horizontal and vertical components of velocity at that moment. vx = axt vx = (2.6 m/sec2) (8.7 sec) vx = 22.6 m/sec vy = ayt vy = (3.4 m/sec2) (8.7 sec) vy = 29.6 m/sec v = (29.62 + 22.62)1/2 v = 37.2 m/sec Ø = tan-1(29.6/22.6) Ø = 52.6°

58*. A particle travels between two parallel walls separated by 15 m. It moves toward the opposing wall at a constant rate of 3.0 m/sec and moves parallel to the walls with an acceleration of 1.7 m/sec². If it starts from rest, what will its velocity be when it hits the opposing wall? (State both its speed and its direction.)

H) and V) will identify the horizontal and vertical equations. H) sx = vxt t = sx/vx t = 15 m / (3 m/sec) t = 5 sec V) vy = ayt vy = (1.7 m/sec2) (5 sec) vy = 8.5 m/sec Ø = tan-1(8.5/3.0) Ø = 70.6° v = (8.52 + 3.02)1/2 v = 9.01 m/sec

In our previous graphs of velocity vs. time we had mostly constant velocity, represented by a horizontal line. Now we will consider velocities which change continually and steadily (i.e.: cases of constant acceleration).

59. On graphs of velocity vs. time, plot the following:
a) Starting at rest, accelerate at 3 m/sec² for 3 seconds.
b) Starting at 8 m/sec, accelerate 2 m/sec² for 3 seconds.

a) At t = 1, v = (3 m/sec2) (1 sec) = 3 m/sec At t = 2, v = (3 m/sec2) (2 sec) = 6 m/sec At t = 3, v = (3 m/sec2) (3 sec) = 9 m/sec

b) At t = 0, v = (2 m/sec2) (0 sec) + 8 m/sec = 8 m/sec At t = 1, v = (2 m/sec2) (1 sec) + 8 m/sec = 10 m/sec At t = 2, v = (2 m/sec2) (2 sec) + 8 m/sec = 12 m/sec At t = 3, v = (2 m/sec2) (3 sec) + 8 m/sec = 14 m/sec

60. Again, on velocity vs. time:
a) Start at 6 m/sec and accelerate at -3 m/sec² for 3 seconds.
b) Start at -5 m/sec, accelerate at 2 m/sec² for 4 seconds.

a)

b)

61. Express graphically the following motions. (plot v vs. t)
a) The particle travels at 3 m/sec for 2 seconds, then accelerates at 5 m/sec² for 3 seconds.
b) With an initial backward motion of 4 m/sec for 3 seconds, the particle then accelerates at 2 m/sec² for 6 seconds.

a)	b) note: In the region with a positive slope the particle is accelerating in a positive direction, but it is moving in a negative direction!! Think about it.

62*. Again, a velocity vs. time graph: the particle starts at -3 m/sec for 4 seconds, then accelerates at 5 m/sec² for 3 seconds, travels at a constant velocity for 2 seconds, and accelerates at -4 m/sec² for 3 seconds.

63* Describe the motion graphed below.

Motion begins at a little over 3 m/sec and begins to accelerate-leveling off at 7.5 m/sec. The object then begins to accelerate in a negative direction, slowing down at -5 m/sec and eventually leveling off at -7 m/sec.

As you sketch graphs of velocity, acceleration, etc., try to imagine what is happening physically. e.g.: a steeper slope means more rapid acceleration, a downward slope means negative acceleration, when the curve crosses the axis the particle has stopped and is now turning around.

Now we return to distance determinations. In each of the following, graph velocity vs. time for the described motion and determine how far it travels.

64. Starting at rest, the particle accelerates at 6 m/sec² for 8 seconds.

Remember, the distance from the start can be found by determined the area enclosed by the v vs. t graph. s =(1/2)bh S = 1/2 (8 sec) (48 m/sec) s = 192 m

65. Starting from rest, 8 m from the origin, it accelerates at 4 m/sec² for 5 seconds.

s = 1/2 (5 sec) (20 m/sec) s = 50 m Were you tricked? the 8 m is irrelevant in determining how far something has traveled, and will not show up on a vs. t graph.

66. Starting at 3 m/sec, -5 m from the origin, it accelerates at 2 m/sec² for 7 seconds.

Break it into two regions: A = (1/2)bh A = 1/2 (7) (14) A = 49 A = bh A = (7) (3) A = 21 s = 49 + 21 s = 70m

67. Starting at 4 m/sec, accelerate at -2 m/sec² for 3 seconds.

To get the final position, the negative area of ii must be added to i. s = 1/2 (4 m/sec) (2 sec) + 1/2 (-2 m/sec) (1 sec) s = 4 - 1 s = 3 m This makes sense, in region ii the motion is backwards.

68. A car traveling at 30 m/sec brakes to a stop in 4 seconds. How far does it travel in this time?

s = 1/2 (4 sec) (30 m/sec) s = 60 m

69*. A rocket travels at a constant rate of 5.9 x l04 cm/sec for 97 seconds. Then it accelerates at 3.1 x l02 cm/sec² for 60 seconds.
a) How fast is it traveling at the end of the time?
b) How far is it from its origin?

a) v = at + v0 v = (3.1 * 102 cm/sec2) (60 sec) + 5.9 * 104 cm/sec v = 7.76 * 104 cm/sec

b) s = vt + vt s = (5.9 * 104) (97) + ((5.9 * 104 + 7.76 * 104) / 2) * 60 s = 9.67 * 106 cm

The following are important generalizations of what we have just been doing. It is essential that the solutions seem reasonable and easy to you after you have studied them for a while.

70. Find the distance traveled by a particle starting at rest and accelerating at rate a for time t. (first graph v vs.t)

At time t the particle is traveling at v = at. The area under the curve, then, is: s = 1/2 (t) (at) s = (1/2)at2

71. Find the distance traveled by a particle starting at velocity vo and accelerating at rate a for time t.

At time t the particle moves at velocity v = at + v0 thus, from the area of a trapezoid: s = ((v0 + at + v0) / 2)t s = (v0 + (1/2)at)t s = (1/2)at2 + v0t

This last equation is of enormous importance to the understanding of motion. It should be memorized as s = 1/2 at2 keeping in mind that when there is an initial velocity, we will add the term vot to get s = 1/2 at2 + vot In the following problems you will not have to resort to cumbersome graphing. Simply apply the equations that have been derived.

72. A particle accelerates from rest at 28 m/sec² for 6.0 seconds. How far does it get?

s = (1/2)at2 s = 1/2 (28 m/sec2) (6.0 sec)2 s = 504 m

73. A Hostess Twinkie accelerates from rest at 6.3 m/sec² for a distance of 35m. How long does this take?

s = (1/2)at2 t = (2s/a)1/2 t = [2(35 m)/(6.3 m/sec2)]1/2 t = 3.3 sec

74. A particle accelerates from rest for 5.4 seconds and travels at 125 m in the process. What is its acceleration?

s = (1/2)at2 a = 2s/t2 a = [(2) (125 m)] / [(5.4 sec)2] a = 8.6 m/sec2

75. A banana starts at 9.0 m/sec and accelerates at 4.2 m/sec² for 3.0 seconds. How far does it get?

This time there is am initial velocity v0, so we may use the equation: s = (1/2)at2 + v0t or s = 1/2 (4.2 m/sec2) (3.0 sec)2 + (9.0 m/sec) (3.0 sec) s = 45.9 m

76. A car traveling at 30 mi/hr accelerates uniformly for 8 seconds, covering 520 ft in this time. What was its acceleration? (Watch your units!)

Convert mi/hr into ft/sec: (30 mi/hr) ((88 ft/sec)/(60 mi/hr)) = 44 ft/sec s = (1/2)at2 + v0t a = 2(s - v0t/t2 a = [(2) (520 - 44 * 8)] / (8 sec)2 a = 5.25 ft/sec2

77. A duck, skidding backward on ice at -5 m/sec, accelerates at +4 m/sec². How long will it take to reach +8 m?

s = (1/2)at2 + v0t (a/2)t2 + v0t - s = 0 (4/2)t2 + (-5)t - 8 = 0 (physically unreasonable) t = (-b + (B2 - 4AC))1/2 / 2A t = (5 + [52 - (4) (2) (-8)]1/2) / 2(2) t = -1.1 sec, 3.6 sec

78. A particle accelerates at -5 m/sec² for 6.7 seconds and ends up going -18.0 m. What was its initial velocity?

s = (1/2)at2 + v0t v0 = (s - (1/2)at2) / t v0 = s/t - (1/2)at v0 = -18/6.7 - 1/2(-5) (6.7) v0 = 14.1 m/sec

79. A cucumber, traveling at 9.7 m/sec accelerates at -1.8 m/sec². How long will it take to get back to the starting point?

When it returns to the starting point, the particle will be at s = 0, SO: s = (1/2)at2 + v0t 0 = 1/2 (-1.8)t2 + 9.7t Solving: 1/2 (1.8)t2 = 9.7t t = 0 t = 10.8 sec

80*. Two particles take off from the same point at the same time. Particle A has an initial velocity of 8.7 m/sec and an acceleration of 3.0 m/sec². Particle B has an initial velocity of 2.3 m/sec and an acceleration of 4.7 m/sec². At what time will B pass A?

sA = (1/2)at2 + v0t sA = 1/2 (3.0)t2 + 8.7t sB = (1/2)at2 + v0t sB = 1/2 (4.7)t2 + 2.3t The particles meet when: sA = sB 1/2 (3)t2 + 8.7t = 1/2(4.7)t2 +2.3t This simplifies to: .85t2 - 6.4t = 0 t = 0 t = 7.53 sec

81*. Particle A starts with a velocity of 11.2 m/sec and an acceleration of 4.3 m/sec². Particle B gets a 35 m head start at 6.1 m/sec with an acceleration of 3.9 m/sec². When do the particles meet?

sA = (1/2)at2 + v0t sA = 1/2 (4.3)t2 + 11.2t sB = (1/2)at2 + v0t sB = 1/2 (3.9)t2 + 6.1t + 35 Equating: 2.15t2 + 11.2t = 1.95t2 + 6.1t +35 or .20t2 + 5.1t - 35 = 0 t = (-5.1 + [5.12 - (4) (.2) (-35)]1/2) / 2 (.2) t = 5.62 sec, -31.1

82*. This time we consider a collision. Particle A starts from rest and accelerates to the right at 64.2 cm/sec². Particle B starts at 372 cm and moves at constant velocity of -48.9 cm/sec. When will they meet?

sA = (1/2)at2 sA = 1/2 (64.2)t2 sB = v0t + s0t sB = -48.9t + 372 (64.2 / 2)t2 = -48.9t + 372 Think about equation sB: 32.1t2 + 48.9t - 372 = 0 Particle b starts at 372 m: t = (-48.9 + [48.92 - (4) (32.1) (-372)]1/2) / 2 (32.1) t = 2.73 sec, -4.25 And gets closer to the origin. Note how the negative velocity reflects this.