 Phyllis Fleming Physics Physics 104

 Answers - One Dimensional Motion
 1 (a) The average speed = distance traveled/time = 100 m/10 s = 10 m/s. (b) time = distance/average speed = 1610 m/10 m/s = 161 s = 161 s x (1 min/60 s) = 2.68 min.
 2 If y = kx, y/x = k = a constant. y is directly proportional to x. (a) If you double x, you double y. (b) If you halve x, you halve y. (c) A plot of y as a function of x yields a straight line through the origin.
 3 If y = kx2, y/x2 = k = a constant. y is directly proportional to x2. (a) If you double x, y goes up by a factor of 4. (b) If you halve x, y is reduced by a factor of four. (c) A plot of y as a function of (i) x is a parabola; (ii) the square of x is a straight line through the origin.
 4 (a) The distance moved x divided by the time t equals the constant velocity v. x/t = v or x = vt = 15 m/s (2.0 s) = 30 m. (b) Since x/t = v = constant, the distance moved is directly proportional to the time elapsed. If you double the time, you double the distance. For t = 4.0 s, x = 2 x 30 m = 60 m.
 5 In general for constant acceleration, the distance moved by the object x = xo + vot + 1/2 at2, where xo is the position of the object at t = 0 (the initial position), vo the velocity at t = 0 (the initial velocity), a the acceleration and t the time at which you wish to find x. Let's arbitrarily take xo = 0. Then for an object initially at rest, vo = 0, x = 0 + 0 + 1/2 at2. (a) x(2.0 s) = 1/2 (10 m/s2)(2.0 s)2 = 20 m. (b) x(4 s) = 1/2 (10 m/s2)(4 s)2 = 80 m. For constant acceleration and an initial position and initial velocity of zero, x = 1/2 at2 or x/t2 = 1/2 a = a/2 = a constant. For this case x is directly proportional to the square of t. If you double t, t2 goes up by a factor of four and x must go up by a factor of 4 = 4 x 20 m = 80 m. (c) For this part, we still take the initial position xo = 0, but now the initial velocity vo = 4.0 m/s. x(t) = 0 + vot + 1/2 at2.      x(2.0 s) = 0 + (4.0 m/s)(2.0 s) + 1/2 (10 m/s2)(2.0 s)2 = 28 m (d) x(4.0 s) = 0 + (4.0 m/s)(4.0 s) + 1/2 (10 m/s2)(4.0 s)2 = 96 m With x = vot + 1/2 at2 and x/t2 = vo/t + a/2. Since t is a variable, the right hand side of the above equation is not equal to a constant and x is no longer proportional to the square of the time, that is, 96 ≠ 4 x 28.
 6 By the definition of constant acceleration, a = (v - vo)/(t - 0), where v is the velocity of the object at time t and vo is the velocity of the object at time t = 0. (a) a = {(60 - 20) m/s}/(2.0 s - 0) = 20 m/s2. (b) Taking xo =0,      x(t) = vot + 1/2 at2.      x(2.0 s) = 20 m/s(2.0 s) + 1/2 (20 m/s2)(2.0 s)2 = 80 m.
 7 Distance traveled = speed x time = 3.0 x 108 m/s x 20/3 s = 20 x 108 m. Room lengths = distance traveled/ "room length" = 20 x 108 m/20 m = 108.
 8 Time to travel distance = distance/speed = 60 ft/132 ft/s = 0.454 s.
 9 (a) The average speed = x/t = (v + vi)/t, where v = 15 m/s and vi is the initial speed. The average speed = x/t = 60 m/6.0 s = (15 m/s + vi)/2 or 10 m/s = (15 m/s + vi)/2 and 20 m/s = (15 m/s + vi) so vi = 5 m/s. (b) The acceleration a = (v - vi)/(t - ti) = (15 m/s - 5 m/s)/(6.0 s) = 5/3 m/s2. (c) vi2 = vo2 + 2a(xi - xo).   (5 m/s)2 = 0 + 2(5/3 m/s2)(xi - xo) 25 m2/s2 = (10/3 m/s2)(xi - xo) or (xi - xo) = 7.5 m. Let's check all of this. First we find the time for the object to acquire a velocity vi = 5.0 m/s with an acceleration of 5/3 m/s2 starting from an initial velocity of zero. v = vo + at or 5.0 m/s = 0 + (5/3 m/s2)t or t = 3.0 s.  At t = (6.0 + 3.0)s, x(9.0 s) = 1/2 (5/3 m/s2)(9.0 s)2 = 67.5 m = 60 m + 7.5 m, where 60 m is the distance given in the statement of the problem for the object to go from 5 m/s to 15 m/s and 7.5 m is the distance moved from the point where the object was at rest to where it was traveling with 5 m/s.
 10 The answers are written on the graph (below) for each interval. Note that in #1 the slope of the x vs t curve is positive and increasing in magnitude so the velocity is positive and increasing and the acceleration is positive. In #3 the velocity is positive, but decreasing in magnitude so the acceleration is negative. In #5 the velocity is negative, but it is increasing in magnitude so the acceleration is negative. 11 (a) For constant acceleration v = vi + at. For the car to come to a stop, the final velocity v = 0. 0 = 80 ft/s - (20 ft/s2)t or t = 4.0 s. (b) In general for constant acceleration x = xo + vit + 1/2 at2. The distance moved = x - xo = vit + 1/2 at2 = 80 ft/s(4.0 s) + 1/2 (-20 ft/s2)(16s 2) = 16 ft. (c) During the (1 1/2 + 1/2) s = 2.0 s before the brake is applied, the car traveling at a speed of 80 ft/s goes 80 ft/s(2.0 s) = 160 m. After the brake is applied the car travels 160 ft. The total distance traveled is (160 + 160)ft = 320 ft. (d) Number of "car lengths" traveled = distance traveled/ "length of car" = 320 ft/15 ft = 21.3.
 12 The magnitude 10 m/s of the particle is the same at t = 1.0 s and t = 5.0 s. If we take to the right to be positive, the velocity is to the right at t = 1.0 s , but to the left at t = 5.0 s. For motion along a straight line, we are free to use a positive sign to indicate motion in one direction and a negative sign to indicate motion in the opposite direction. We usually take to the right as positive and to the left as negative.
 13 See Figure 2 below: The acceleration at any instant equals the slope of v versus t at that instant. From t = 0 to t = 1.0 s, a = (15 - 5)m/s/(1 - 0 )s = 10 m/s2. From t = 1.0 s to 2.0 s, the velocity is constant and the acceleration a = 0. From t = 2.0 s to 3.0 s, a = (0 - 15.0)m/s/(3.0 - 2.0)s = -15 m/s2. From t = 3.0 s to 4.0 s, the velocity equals zero and the acceleration a = 0. From t = 4.0 s to 5.0 s, a = (-15.0 - 0)m/s/(5.0 -4.0)s = -15 m/s2. The distance moved by an object in time t equals the area under the v vs t curve from t = 0 to t. From t = 0 to t = 0.5 s, area equals the area of a triangle of height 5.0 m/s and base 0.5 s and square with sides= 5.0 m/s and 0.5 s. Area = (1/2)(5.0 m/s x 0.5 s) + (5 m/s x 0.5s) = 3.75 m. From t = 0.5 s to t = 1.0 s, area = (1/2 x 5.0 m/s x 0.5 s) + (10 m/s x 0.5 s)= 6.25 m. From t = 0 to t = 1, x(1s) = (3.75 + 6.25)m =10 m. From t = 1 to t = 2 s, area = 15 m/s x 1 s = 15 m. x(2s) = (10 + 15) m = 25 m. From t = 2 s to t = 3 s, area = 1/2(15 m/s)(1s) = 7.5 m. x(3s) = (25 + 7.5)m = 32.5 m. From t = 3 to 4 s, the area = 0. x(4s) = 32.5 m. From t = 4 to 5 s, area = 1/2(-15 m/s)(1 s) = -7.5 m. x(5s) = (32.5 - 7.5)m = 25 m. The slope of x vs t at any instant is the velocity at that instant. From t = 0 to t = 1 s, the slope and velocity are positive and increasing. From t = 1 to t = 2 s, the slope and velocity are constant and positive. From t = 2 to t = 3 s, the slope and velocity are positive, but decreasing. From t = 3 to t = 4 s, the slope and velocity equal zero. From t = 4 to t = 5 s, slope and velocity are negative, but increasing. All agree with v versus t in Fig. 2a. 14 Three equations are used frequently for motion with constant acceleration: For motion in the y-direction, (1) v(t) = vo + at (2) y(t) = yo + vot + 1/2 at2, and (3) v2(y) = vo2 + 2a(y - yo). The first is just the definition of constant acceleration, a = (v - vo)/(t - 0), rearranged algebraically. Eq. 1 gives the velocity of an object as a function of time. If you know the initial velocity vo and the acceleration a of the object, you can find its velocity v at time t. The second equation is derived from the area under the velocity versus time curve and the definition of acceleration. It gives the position y as a function of t. If you know the initial position yo of the object, vo and a, you can find y for any time t. The third equation is found by solving for t in Eq. (1) and substituting it into Eq. (2). It gives the velocity v as a function of height y. If you know yo, vo, and a you can find y for any value of v. (a) We know the height reached y, the time t for it to reach this height, the acceleration a of the ball, and taking yo = 0, we can find the one unknown vo with           40 m = 0 + vo(2.0 s) - 1/2 (10 m/s2)(4.0 s2)           40 m = vo(2.0 s) - 20 m or vo = 30 m/s (b) v(2.0s) = vo + at = 30 m/s - 10 m/s2(2 s) = 10 m/s (c) We must interpret the meaning of "how much higher will the ball go." The answer, of course, is "until it stops rising." At the highest point the ball comes momentarily to rest and v(t) = 0. 0 = v2 = vo2 + 2a(y - yo) = (30 m/s)2 - 20 m/s2(y - 0),  or 0 = 900 m2 - 20 m(y) and y = 45 m. Added height = (45 - 40)m = 5 m.
 15 (a) We do not know the time, but taking yo = 60 m and y = 0, we can use v2(y) = vo2 + 2ay(y - yo) v2(0) = (-20 m/s)2 - 20 m/s2(0 - 60 m)          = (400 + 1200)m2/s2   and       v = - 40 m/s, where the negative sign occurs because we have taken up as positive. (b) v(t) = vo + at or -40 m/s = -20 m/s - 10 m/s2 t and t = 2 s. (c) Now the initial and final positions are the same, but the initial velocity vo = +20 m/s. The algebra, however, will be the same as in part (a) and the answer is again v = -40 m/s when it hits the ground. The time does change with v(t) = vo + at and vo = +20 m/s, -40 m/s = +20 m/s -10 m/s2 t and now t = 6 s.
 16 (a) "Coming to rest in a distance of 20 m" means x(t) - xo = 20 m. v2(x) = vo2 + 2a(x - xo). 0 = (10 m/s)2 + 2a(20 m), or(-40a) m = 100 m2/s2, or a = -2.5 m/s2. (b) v(t) = vo + at 0 = 10 m/s - (2.5 m/s2)t.   t = 4.0 s. (c) and (d) are shown in the figure below. 17 The average velocity = 2.0m/0.10s = 20 m/s» v at top or bottom of window. v2 = vo2 + 2a(y - yo) or (20 m/s)2 = 0 + 2(-10 m/s2)(0 - yo) and yo = 20 m. Since the distance between floors is 4.0 m, it fell from an apartment 5 floors up or since she is on the fifth floor, from an apartment on the tenth floor.
 18 For the ball thrown downward, y(t) = yo + vot + 1/2 at2  or taking up as +, y(t) = 40 m - (8.0 m/s)t - (5.0 m/s2)t2      (Equation 1) For the ball thrown upward from the ground, y(t) = 0 + (12 m/s)t - (5.0 m/s2)t2           (Equation 2) When the balls collide the two y's are equal: 40 m - (8.0 m/s)t - (5.0 m/s2)t2 =                           0 + (12 m/s)t - (5.0 m/s2)t2  or 40 m - (8.0 m/s)t = (12 m/s)t 40 m = (20 m/s)t  and  t = 2.0 s. From Eq. 2, y(2.0s) = (12 m/s)(2.0 s) - (5.0 m/s2)(2.0s)2 = 4.0 m From Eq. 1, y(t) = 40 m - (8.0 m/s)(2.0 s) - (5.0 m/s2)(2.0 s)2= 4.0 m. In general, v(t) = vo+ at. For the second ball, v(2.0 s) = 12 m/s - (10 m/s2)(2.0 s) = -8.0 m/s. Since the velocity at 2.0 s of the second ball is negative, it is on its way down when the first ball collided with it.
 19 (a) In Fig. 19a below, at t = 1 s, the velocity is zero (the slope of x vs t at t = 1 s is zero). The acceleration is positive because the slope of x versus t after t = 1 s is positive and increasing. For example, imagine the slope of x vs t at t = 2 s is 1 cm/s, then a = [(1 - 0)cm/s]/(2 - 1) s = +1/2 cm/s2. (b) In Fig. 19b, at t = 1 s, the velocity is zero (the slope of x vs t at t = 1 s 1s zero). For t > 1s, the acceleration is negative because the slope is less than zero and increasing. For example imagine the slope of x vs t at t = 2 s is -1 m/s, then a = [(-1 - 0)cm/s]/(2 - 1)s = -1/2 cm/s2. (c) In Fig. 19c, at t = 1 s, the velocity is negative (the slope of x vs t at t = 1 s is negative). The acceleration is positive because the slope of x vs t after t = 1 s is negative and decreasing. For example imagine the slope of x vs t at t = 1 s is -1 cm/s and at t = 2 s it is 0. Then a = {0 - (-1)cm/s}/(2 - 1)s = 1/2 cm/s2. (d) In Fig. #19d, at t = 1 s, the velocity is negative (the slope of x vs t at t = 1 s is negative). The acceleration is negative because the slope of x vs t is negative and increasing. For example, imagine at t = 1 s,v = -3 m/s and at t = 2s, v = -4 m/s, a = {(-4) - (-3)}cm/s/(2 - 1)s= -1/2 cm/s2. (e) The speed, the magnitude of the velocity, at t = 1 s is increasing for all cases except (c). Homepage Sitemap
 Website Designed By: Questions, Comments To: Date Created: Date Last Updated: Susan D. Kunk Phyllis J. Fleming September 25, 2002 April 1, 2003