Homework## Ch 2, Motion in One Dimension

Ch 2: 4, 9, 12, 14, 19, 21, 29, 39, 40

Questions 4, 5, 6, 9, 16

| Homework Assignment Page | PHY 1350's Home Page | Hmwk, Ch 3 |

Additional problems from Serway's fourth edition

(4 ed) 2.1The position-time graph for a particle moving along the z-axis is as shown in an old Figure P2.1. Determine whether the velocity is positive, negative, or zero at times t_{1}, t_{2}, t_{3}, and t_{4}

(4 ed) 2.2The new BMWM3can accelerate from zero to 60 mi/h in 5.6 s

(a)What is the resulting acceleration in m/s^{2}?

(b)How long would it take the BMW to go from 60 mi/h to 130 mi/hat this rate?

(4 ed) 2.3A hot air balloon is traveling vertically upward at a constant speed of 5.00 m/s. When it is 21.0 m above the ground, a package is released from the balloon.

(a)After it is released, for how long is the package in the air?

(b)What is its velocity just before impact with the ground!

(c)Repeat(a)and(b)for the case of the baloondescendingat 5.0 m/s.

(4 ed) 2.4A hockey player is standing on his skates on a frozen pond when an opposing player skates by with the puck, moving with a uniform speed of 12.0 m/s. After 3.00 s, the first player makes up his mind to chase his opponent. If the first player accelerates uniformly at 4.00 m/s^{2},

(a)how long does it take him to catch the opponent?

(b)How far has the first player traveled in this time?

Conceptual Questions

2.Q4 Is it possible to have a situation in which the velocity and acceleration have opposite signs? If so, sketch a velocity-time graph to prove your point.Certainly. Consider a car moving to the right but slowing down. Moving to the right means its velocity is positive. Slowing down means its velocity is

decreasing or thechangein velocity is negative and that means the acceleration is negative.

2.Q5 If the velocity of a particle is nonzero, can its acceleration be zero? Explain.If the velocity is

constantthe acceleration iszero.

2.Q6 If the velocity of a particle is zero, can its acceleration be nonzero? Explain.A velocity of

zerois also aconstantvelocity and that means the acceleration iszero.

2.Q9 A student at the top of a building of height h throws one ball upward with an initial speed v_{yi}and then throws a second ball downward wit the same initial speed. How do the final speeds of the balls compare when they reach the ground?This one will be fun or interesting to talk about again after we have studied Energy Conservation.

When the ball that was initially thrown

upward comes back to the height of the top of the building, its speed is again v_{yi}. It will have the samespeed. Of course itsvelocitywill be - v_{yi}because it is movingdown. That means that it has exactly the same speed and velocity at the ball that is initially throwndownward with initial speed v_{yi}so the two balls hit the ground with exactly the same speed (and velocity!).

2.Q16 A pebble is dropped into a water well and the splash is heard sixteen seconds later; as illustrated in the "BC" cartoon in Figure !2.16. Estimate the distance from the top of the well to the water's surface.We can use the equation

y = y

_{i}+ v_{yi}+ (1/2) a_{y}t^{2}We are measuring distances from the top of the well so y

_{i}= 0 and wedropthe pebble so v_{yi}= 0 so this equation reduces toy = (1/2) a

_{y}t^{2}y = (1/2) ( - 9.8 m/s

^{2}) (16 s)^{2}y = - 1 254.4 m

That is a very deep "well", indeed! But, maybe, wells were deeper in such prehistoric times. If we had used the approximation that a

_{y}= - g = - 10 m/s^{2}, then our value would have beeny = - 1 280 m

Problems from the current (5th) edition of Serway and Beichner.

2.4A particle moves according to the equation x = 10 t^{2}where x is in meters and t is in seconds.

(a)Find the average velocity for the time interval from 2.0 s to 3.0 s.x(2 s) = 10 (2)^{2}= 10 (4) = 40 mx(3 s) = 10 (3)

^{2}= 10 (9) = 90 mx = x

_{f}- x_{i}= x(3 s) - x(2 s) = 90 m - 40 m = 50 mt = 1.0 s

v

_{avg}= x / t =^{50 m}/_{1.0 s}= 50^{m}/_{s}

(b)Find the average velocity for the time interval 2.0 s to 2.1 sx(2 s) = 10 (2)^{2}= 10 (4) = 40 mx(2.1 s) = 10 (2.1)

^{2}= 10 (4.41) = 44.1 mx = x

_{f}- x_{i}= x(2.1 s) - x(2 s) = 44.1 m - 40 m = 4.1 mt = 0.1 s

v

_{avg}= x / t =^{4.1 m}/_{0.1 s}= 41^{m}/_{s}[[

(c)Find the instantaneous velocity at t = 2.0 sv = dx/dt =^{dx}/_{dt}= 10 [ d(t^{2})/dt ] = 10 [ 2t ] = 20 tv(2 s) = 20 (2) = 40 m/s = 40

^{m}/_{s}_{ ]]}

2.9The position-time graph for a particle moving along the x-axis is shown in Figure P2.9

(a)Find the average velocity in the time interval t = 1.5 s to t = 4.0 s. From the graph, we can findx (1.5 s) = 8 mand

x(4.0 s) = 2 mx = x

_{f}- x_{i}= x(4.0 s) - x(1.5 s) = 2 m - 8 m = - 6 mt = t

_{f}- t_{i}= 4.0 s - 1.5 s = 2.5 sv

_{avg}= x / t =^{- 6 m}/_{2.5 s}= - 24^{m}/_{s}Remember, anything means (final value) - (initial value)

(b)Determine the instantaneous velocity at t = 2.0 s by measuring the slope of the tangent line shown in the graph.To measure the slope of the line drawn tangent to the curve at t = 2.0 s, we can picktwo pointson the line.We might choose

(t = 1.0 s, x = 9.0 m)and

(t = 3.5 s, x = 1.0 m)Now we calculate v = x / t

x = x_{f}- x_{i}= 1.0 m - 9.0 m = - 8.0 mt = t

_{f}- t_{i}= 3.5 s - 1.0 s = 2.5 sv

_{avg}= x / t =^{- 8.0 m}/_{2.5 s}= - 32^{m}/_{s}

(c)At what value of t is the velocity zero?The velocity is zero when the slope of the tangent line (on an x-, t- graph) is zero.That happens on this graph for

t = 4.0 s.

2.12A particle is moving with a velocity v_{o}= 60.0 m/s in the positive x direction at t = 0. Between t = 0 and t = 15 s, the velocity decreases uniformly to zero. What is the average acceleration during this 15-s interval. What is the significance of thesignin your answer?a = v / tv = v

_{f}- v_{i}= v_{f}- v_{o}= 0 - 60^{m}/_{s}= - 60^{m}/_{s}t = 15 s

a = v / t = [ - 60

^{m}/_{s}] / [ 15 s ] = - 4 (m/s)/s = - 4 m/s^{2}The minus sign means the particle is

slowingdown. Its velocity isdecreasing. It isdecelerating.

2.14A particle starts from rest and accelerates as shown in Figure P2.14.Determine the following:

(a)the particle's speed at t = 10 s and at t = 20 s andFrom the graph, we can see that from t = 0 to t = 10 s, the acceleration is a constant 2.0 m/s^{2}v = v_{i}+ a tv

_{i}= 0a = 2 m/s

^{2}v(10 s) = 0 + (2 m/s

^{2}) (10 s)

v(10 s) = 20 m/sFor the next five seconds, from t = 10 s to t = 15 s, the acceleration is

zerowhich means the velocity remainsconstant,v(15 s) = 20 m/sFor the next five seconds, from t = 15 s to t = 20 s, the acceleration is negative, a = - 3.0 m/s

^{2}v = v_{i}+ a tv

_{i}= v(15 s) = 20 m/sa = - 3 m/s

^{2}v(20 s) = 20 m/s + (- 3 m/s

^{2}) (5 s)

v(20 s) = 5 m/s

(b)the distance traveled in the first 20 sWe have already determined the values of the acceleration for various time intervalsx = x_{i}+ v_{i}t + (^{1}/_{2}) a t^{2}x

_{i}= 0v

_{i}= 0a = 2 m/s

^{2}x(10 s) = 0 + 0 + (

^{1}/_{2}) (2 m/s^{2})(10 s)^{2}

x(10 s) = 100 mThe acceleration remains

constant(at a = 0) for the next five seconds (until t = 15 s) so we can, again, apply this equation which describes distance with constant acceleration.x = x_{i}+ v_{i}t + (^{1}/_{2}) a t^{2}x

_{i}= x(10 s) = 100 mv

_{i}= v(10 s) = 20 m/sa = 0

x(15 s) = 100 m + (20 m/s)(5 s) + (

^{1}/_{2}) (0)(5 s)^{2}x(10 s) = 200 m

The acceleration again remains

constant(at a = - 3 m/s^{2}) for the next five seconds (until t = 20 s) so we can, again, apply this equation which describes distance with constant acceleration.x = x_{i}+ v_{i}t + (^{1}/_{2}) a t^{2}x

_{i}= x(15 s) = 200 mv

_{i}= v(15 s) = 20 m/sa = - 3 m/s

^{2}x(20 s) = 200 m + (20 m/s)(5 s) + (

^{1}/_{2}) ( - 3 m/s^{2})(5 s)^{2}

x(20 s) = 262.5 m

2.19Figure P2.19 shows a graph of v versus t for the motion of a motocyclist as she starts from rest and moves along the road in a straight line.

(a)Find the average acceleration for the time interval t_{o}= 0 to t_{1}= 6.0 s.From the graph, we can read the velocities for these timesv_{1}= v(t_{1}) = v(6 s) = 8 m/sv

_{o}= v(t_{o}) = v(0 s) = 0v = v

_{1}- v_{o}= 8 m/s - 0 = 8 m/st = 6.0 s

a = v / t = [8 m/s ] / [6 s] = 1.33 m/s

^{2}

a = 1.33 m/s^{2}

(b)Estimate the time at which the acceleration has its greatest positive value and the value of the acceleration at this instant.Acceleration is theslopeof the line on a velocity-time graph like Figure P2.23. The slope seems greatest at aboutt = 3 sAnd, there, I estimate the slope by sketching a tangent line which goes through the points

(t = 1 s, v = 0) and (t = 6 s, v = 10 m/s)v = v

_{f}- v_{i}= 10 m/s - 0 = 10 m/st = 6.0 s

a = v / t = [10 m/s ] / [6 s] = 1.67 m/s

^{2}

a = 1.67 m/s^{2}

(c)When is the acceleration zero?Zero acceleration means the slope of the line tangent to the curve on a v-t graph iszero.a = 0 fort = 6 s

(d)Estimate the maximumnegativevalue of the acceleration and the time at which it occurs.Acceleration is theslopeof the line on a velocity-time graph like Figure P2.23. The slope isnegativefrom t = 0 to about t = 10 seconds. During that time, the slope seems greatest (negatively) at about t = 8 s. T here, I estimate the slope by sketching a tangent line which goes through the points(t = 6 s, v = 10 m/s) and (t = 11 s, v = 0 m/s)v = v

_{f}- v_{i}= 0 m/s - 10 m/s = - 10 m/st = 5.0 s

a = v / t = [- 10 m/s ] / [5 s] = - 1.8 m/s

^{2}

a = - 1.8 m/s^{2}

2.21Jules Verne in 1865 proposed sending people to the Moon by firing a space capsule from a 220-m-long cannon with a final velocity of 10.97 km/s. What would have been the unrealistically large acceleration experienced by the space travelers during launch? Compare your answer with the free-fall acceleration of 9.8 m/s^{2}.First, let's change the final velocity to units of m/s; we can almost do that in our heads.

v

_{f}= 10.97^{km}/_{s}[^{1000 m}/_{km}] = 10.97 x 10^{3}^{m}/_{s}

2.29A drag racer starts her car from rest and accelerates at 10.0 m/s^{2}for the entire distance of 400 m (^{1}/_{4}mile).

(a)How long did it take the car to travel this distance?x = x_{i}+ v_{i}t + (^{1}/_{2}) a t^{2}400 m = 0 + 0 + (

^{1}/_{2}) (10 m/s^{2}) t^{2}t

^{2}= 80 s^{2}

t = 8.94 s

(b)What is its speed at the end of the run?v = v_{i}+ a tv = 0 + (10 m/s

^{2})(8.94 s)

v = 89.4 m/sv = 89.4 m/s [

^{3600 s}/_{h}] [^{km}/_{1000 m}] = 322^{km}/_{h}v = 322

^{km}/_{h}[^{mi}/_{1.61 km}] = 200^{mi}/_{h}

2.39A ball accelerates at 0.5 m/s^{2}while moving down an inclined plane 9.0 m long. When it reaches the bottom, the ball rolls up another plane, where, after moving 15 m, it comes to rest.

(a)What is the speed of the ball at the bottom of the first plane?v^{2}= v_{i}^{2}+ 2 a (x - x_{i})v

^{2}= 0^{2}+ 2 (0.5 m/s^{2})(9 m) = 9 m^{2}/s^{2}

v = 3 m/s

(b)How long doe is take to roll down the first plane?v = v_{i}+ a t3 m/s = 0 + (0.5 m/s

^{2}) t

t = 6 s

(c)What is the acceleration along the second plane?v^{2}= v_{i}^{2}+ 2 a (x - x_{i})What value of a brings the ball to rest, having

v = 0when it started with

v_{i}= 3 m/safter traveling a distance of

x - x_{i}= 15 m

v^{2}= v_{i}^{2}+ 2 a (x - x_{i})0

^{2}= (3 m/s)^{2}+ 2 a (15 m)

a = - 0.3 m/s^{2}

(d)What is the ball's speed 8.0 m along the second plane?v^{2}= v_{i}^{2}+ 2 a (x - x_{i})v

^{2}= (3 m/s)^{2}+ 2 ( - 0.3 m/s^{2}) ( 8 m)v

^{2}= ( 9 - 4.8 ) (m^{2}/s^{2}) = 4.2 m^{2}/s^{2}

v = 2.05 m/s

2.40Speedy Sue driving at 30 m/s enters a one-lane tunnel. She then observes a slow-moving van 155 m ahead traveling at 5.0 m/s. Sue applies her brakes but can decelerate only at 2.0 m/s^{2}because the road is wet. Will there be a collision?If yes, determine how far into the tunnel and at what time the collision occurs.

If no, determine the distance of closest approache between Sue's car and the van.

We will measure distances from the tunnel's entrance.The position of the

van is given byx_{van}= x_{v}= x_{vi}+ v_{vi}t + (^{1}/_{2}) a_{v}t^{2}x

_{vi}= 155 mv

_{vi}= 5.0 m/sa

_{v}= 0x

_{v}= 155 m + (5.0 m/s) t + 0x

_{v}= 155 m + (5.0 m/s) tThe position of Speedy Sue's

car is given byx_{car}= x_{c}= x_{ci}+ v_{ci}t + (^{1}/_{2}) a_{c}t^{2}x

_{ci}= 0v

_{ci}= 30.0 m/sa

_{c}= - 2.0 m/s^{2}x

_{c}= 0 + (30 m/s) t + (^{1}/_{2}) (- 2.0 m/s^{2}) t^{2}x

_{c}= (30 m/s) t + (^{1}/_{2}) (- 2.0 m/s^{2}) t^{2}To determine if there is a collision, we can set these two positions equal to each other (as they will be if there is a collision) and solve for t, the time at which that collision occurs.

x_{c}= x_{v}(30 m/s) t + (

^{1}/_{2}) (- 2.0 m/s^{2}) t^{2}= 155 m + (5.0 m/s) t(1 m/s

^{2}) t^{2}+ ( - 25 m/s) t + 155 m = 0This is now in the "standard form" of a quadratic equation,

a x^{2}+ b x + c = 0with

x = ta = 1 m/s

^{2}b = - 25 m/s

c = 155 m

So we can now use the quadratic equation to solve for t. We can explicitly keep the units or we can ensure that the units are consistent and simply write

a = 1, b = - 25, and c = 155

t = 11.4 secondsor t = 13.6 seconds

The collision occurs 11.4 seconds after entering the tunnel.Mathematically, t = 13.6 s isalsoa solution, but the collision has already occured about two seconds before that!

Solutions to the additional problems from Serway's fourth edition

(4 ed) 2.1The position-time graph for a particle moving along the z-axis is as shown in an old Figure P2.14. Determine whether the velocity is positive, negative, or zero at times

a)t_{1}; the velocity, as the slope of the tangent line, iszero

b)t_{2}; the velocity, as the slope of the tangent line, isnegative

c)t_{3}; the velocity, as the slope of the tangent line, ispositive

d)t_{4}; the velocity, as the slope of the tangent line, iszero

(4 ed) 2.2The new BMWM3can accelerate from zero to 60 mi/h in 5.6 s

(a)What is the resulting acceleration in m/s^{2}?a = v / t = [ 60 mi/h ] / 5.6 s = 10.7 (mi/h)/s

a = 10.7 mi/h/sa = 10.7

^{mi}/(_{h - s}) [^{1.61 km}/_{mi}] [^{1000 m}/_{km}] [^{h}/_{3600 s}] = 4.79 m/s^{2}

a = 4.79 m/s^{2}

(b)How long would it take the BMW to go from 60 mi/h to 130 mi/hat this rate?v = v_{i}+ a ta t = v - v

_{i}t = [v - v

_{i}] / at = [ 130 mi/h - 60 mi/h ] / (10.7 mi/h/s )

t = [ 70 mi/h ] / (10.7 mi/h/s )

t = 6.54 s

That is 6.45 seconds

beyondthe 5.6 s required to reach 60 mi/h. Thetotal timewill bet_{tot}= 5.6 s + 6.54 s

t_{tot}= 12.1 s

(4 ed) 2.3A hot air balloon is traveling vertically upward at a constant speed of 5.00 m/s. When it is 21.0 m above the ground, a package is released from the balloon.

(a)After it is released, for how long is the package in the air?Once released, the package is in free fall with an acceleration ofa = - g = - 9.8 m/s^{2}We know its initial velocity and initial position

v_{i}= 5 m/s

y_{i }= 21 mThe later position of the package is given by

y = y_{i}+ v_{i}t + (^{1}/_{2}) a t^{2}y = 21 m + (5 m/s) t + (

^{1}/_{2}) ( - 9.8 m/s^{2}) t^{2}Now we set y = 0 and solve for t

y = 0 = 21 m + (5 m/s) t + (^{1}/_{2}) ( - 9.8 m/s^{2}) t^{2}As before, we can carry the units explicitly or we can ensure that we have consistent units and drop them and write only

4.9 t^{2}- 5 t - 21 = 0From the quadradic equation, we find the two solutions of t

t_{1}= 2.64 s, and t_{2}= - 1.62 sPhysically, we are only interested in solutions for t > 0. Mathematically, our equation is only valid for t > 0 since it is valid only

afterthe package is released. So we use only t_{1}.t = 2.64 s

(b)What is its velocity just before impact with the ground!Once released, the velocity of the package is given byv = v_{i}+ a tv = 5 m/s + ( - 9.8 m/s

^{2}) tv = 5 m/s + ( - 9.8 m/s

^{2}) (2.64 s)

v = - 20.9 m/sOf course, the minus sign indicates that the velocity is directed

downward.

(c)Repeat(a)and(b)for the case of the baloondescendingat 5.0 m/s.As before, once released, the package is in free fall with an acceleration ofa = - g = - 9.8 m/s^{2}We know its initial velocity and initial position

v_{i}= - 5 m/sThe package, along with the balloon, is now moving

downwardand this shows up as the negative sign on the velocityy_{i }= 21 mThe later position of the package is given by

y = y_{i}+ v_{i}t + (^{1}/_{2}) a t^{2}y = 21 m + ( - 5 m/s) t + (

^{1}/_{2}) ( - 9.8 m/s^{2}) t^{2}Now we set y = 0 and solve for t

y = 0 = 21 m + ( - 5 m/s) t + (^{1}/_{2}) ( - 9.8 m/s^{2}) t^{2}As before, we can carry the units explicitly or we can ensure that we have consistent units and drop them and write only

4.9 t^{2}+ 5 t - 21 = 0From the quadradic equation, we find the two solutions of t

t_{1}= 1.62 s, and t_{2}= - 2.64 sPhysically, we are only interested in solutions for t > 0. Mathematically, our equation is only valid for t > 0 since it is valid only

afterthe package is released. So we use only t_{1}.t = 1.62 sOnce released, the velocity of the package is given by

v = v_{i}+ a tv = - 5 m/s + ( - 9.8 m/s

^{2}) tv = - 5 m/s + ( - 9.8 m/s

^{2}) (1.62 s)

v = - 20.9 m/sNotice that the velocities are the same. Later on, we can describe this in terms of energy conservation. The

kinetic energyof the package is the same whether it is thrownupwith v = + 5 m/s or if it is throwndownwith v = - 5 m/s.

(4 ed) 2.4A hockey player is standing on his skates on a frozen pond when an opposing player skates by with the puck, moving with a uniform speed of 12.0 m/s. After 3.00 s, the first player makes up his mind to chase his opponent. If the first player accelerates uniformly at 4.00 m/s^{2},

(a)how long does it take him to catch the opponent?We know the initial velocities and accelerations of the two players,a_{1}= 4 m/s^{2}, a_{2}= 0v

_{1i}= 0, v_{2i}= 12 m/sx

_{1i}= x_{2i}= 0The position of player #2 is given by

x_{2}= x_{2i}+ v_{2i}t + (^{1}/_{2}) a_{2}t^{2}x

_{2}= (12 m/s) tBe careful with the

time. We must account for player #1's wait of 3 s. With this accounted for, we can calculate the position of player #1 fromx_{1}= x_{1i}+ v_{1i}(t - 3 s) + (^{1}/_{2}) a_{1}(t - 3 s)^{2}Of course, this equation only makes sense for t > 3 s.

x_{1}= 0 + 0 + (^{1}/_{2}) (4 m/s^{2}) (t - 3 s)^{2}= (2 m/s^{2}) (t^{2}- 6 s t + 9 s^{2})Now we set x

_{1}= x_{2}and solve for the time t.x_{1}= x_{2}(2 m/s

^{2}) (t^{2}- 6 s t + 9 s^{2}) = (12 m/s) tWe can either keep the units in explicitly or ensure that we have consistent units and simply write

2 (t^{2}- 6t + 9) = 12t2 t

^{2}- 12 t + 18 = 12 t2 t

^{2}- 24 t + 18 = 0t

^{2}- 12 t + 9 = 0There are two solutions to this quadratic equation,

t_{1}= 11.2 s, and t_{2}= 0.8 sHowever, the equation for the position of player #1 is not valid for t

_{2}< 3 s, so we keep only t_{x},t = 11.2 s

(b)How far has the first player traveled in this time?Nowwhereis player #2 (and, therefore, player #1 as well) at this time?x_{2}= x_{2i}+ v_{2i}t + (^{1}/_{2}) a_{2}t^{2}x

_{2}= (12 m/s) tx

_{2}= (12 m/s) (11.2 s)x

_{2}= 134.4 m(Is a hockey rink that large?)

| Homework Assignment Page | PHY 1350's Home Page | Hmwk, Ch 3 | (c) Doug Davis, 2001; all rights reserved