Problem 1: For the following measurements, determine the uncertainty implied by the number of significant figures.

Part (a) r = 1.69 m MultipleChoice : 1) ∆r = ± 0.50 m 2) ∆r = ± 0.100 m 3) ∆r = ± 0.10 m 4) ∆r = ± 1 m 5) ∆r = ± 0.1 m 6) ∆r = ± 1.0 m 7) ∆r = ± 1.00 m 8) ∆r = ± 0.5 m 9) ∆r = ± 1.000 m 10) ∆r = ± 0.500 m 11) ∆r = ± 0.01 m

Solution: The true value of r should be in the range 1.69 m 0.005 m. r= 0.005 m is not a choice, so relax the uncertainty and slightly make it bigger to r= 0.01 m.

Part (b) a = -1.7 m/s2 MultipleChoice : 1) ∆a = ± 0.50 m/s2 2) ∆a = ± 0.100 m/s2 3) ∆a = ± 0.10 m/s2 4) ∆a = ± 1 m/s2 5) ∆a = ± 0.01 m/s2 6) ∆a = ± 1.0 m/s2 7) ∆a = ± 1.00 m/s2 8) ∆a = ± 0.5 m/s2 9) ∆a = ± 1.000 m/s2 10) ∆a = ± 0.500 m/s2 11) ∆a = ± 0.1 m/s2

Solution: The true value of a should be in the range -1.7 m/s2 0.05 m/s2. a= 0.05 m/s2 is not a choice, so relax the uncertainty and slightly make it bigger to r= 0.1 m/s2.

Note: It does not make sense for uncertainty to be very precise (otherwise it is not uncertainty) and have many significant figure. So any uncertainty should have only one or two significant figures. Choices like 2), 7), 9), and 10) look funny because they have many significant figures. Part (c) m = 0.241 kg MultipleChoice : 1) ∆m = ±0.001 2) ∆m = 0 3) ∆m = 0.001 kg 4) ∆m = ±0.001 kg Solution: The true value of m should be in the range 0.241 kg 0.0005 kg. m= 0.0005 kg is not a choice, so relax the uncertainty and slightly make it bigger to m= 0.001 kg.

Part (d) t = 0.033 s MultipleChoice : 1) ∆t = 0.001 s

2) ∆t = 0 3) ∆t = ± 0.001 4) ∆t = ± 0.001 s Solution: The true value of a should be in the range 0.033 m/s2 0.0005 s. t= 0.0005 s is not a choice, so relax the uncertainty and slightly make it bigger to t= 0.001 s.

Part (e) v = 210 m/s MultipleChoice : 1) ∆v = ± 1 m/s 2) ∆v = ± 100 m/s 3) ∆v = ± 10 4) ∆v = 10 m/s 5) ∆v = ± 100 6) ∆v = ± 10 m/s

Solution: The true value of v should be in the range 210 m/s2 5 m/s. v= 5 m/s is not a choice, so relax the uncertainty and slightly make it bigger to v=10 m/s. 

   

Problem 2: Report the answer to the following computations with the appropriate number of significant figures. 

Part (a) How would you report d = 8.20000 mm to 3 significant digits? MultipleChoice : 1) d = 8.200 mm 2) All of the answer choices above have the same number of significant digits. 3) d = 8.2000 mm 4) d = 8.20000 mm 5) d = 8.2 mm 6) d = 8.20 mm Solution: 3 significant figures d=8.20 mm

Part (b) (15 kg)•(9.81 m/s2) = ? MultipleChoice : 1) 147 kg•m/s2 2) 100 kg•m/s2 3) 147.15 kg•m/s2 4) 200 kg•m/s2 5) 140 kg•m/s2 6) 150 kg•m/s2

Solution: (15 kg)•(9.81 m/s2) = 147.15 kg•m/s2, but 15kg has only two significant figures so the product is accurate only to two significant figures also. Rounding off 147.15 kg•m/s2 to two significant figures we will get 150 kg•m/s2. Part (c) 15.6 cm - 2.6 mm = ? MultipleChoice : 1) 15.30 cm 2) 10 cm 3) 15.300 cm 4) 20 cm 5) 15 cm 6) 15.3 cm

Solution: 15.6 cm - 2.6 mm = 15.6 cm – 0.26 cm =15.34cm, but 15.6 cm is accurate only up to the first place after decimal, so 15.3cm should be rounded off to 15.3cm.

Part (d) 420 L – 84 L = ? MultipleChoice : 1) 300 L 2) 336 L 3) 400 L 4) 330 L 5) 340 L

Solution: 420L – 84L = 336L, but 420L has only two significant figures,336L should be rounded off to 340L. Part (e) 52 cm - 13.1 cm = ? (Choose all correct answers.) MultipleSelect :

1) 30 cm 2) 40 cm 3) 223 cm 4) 391 mm 5) 39 cm 6) 223 mm 7) 400 mm 8) 39.1 cm 9) 390 mm Solution: 52cm – 13.1 cm = 38.9 cm , but 52 cm is accurate only up to the decimal, so 38.9 cm should be rounded off to 39cm or 390mm.

 

   

Problem 3: Answer the following questions using the following table of base quantities, derived quantities and units. Use dimensional analysis to guide your thinking.

 

 

 

 

Part (a) Using dimensional analysis, determine the units of the quantity ∆t. Select the single best answer. MultipleChoice : 1) m/s 2) no units 3) 1/s2 4) 1/s 5) s2 6) s 7) m/s2

Solution: t = tf – ti .  All three terms t, tf , and ti in this equation must have the same units.  Unit for tf , 

and ti is s, so the unit for t should also be t. 

Part (b) Which of the following will have units of meters? Select all that apply. MultipleSelect : 1) vr 2) v2 3) v/∆t 4) a∆t 5) 1/2a(∆t)2 6) v∆t 7) 2ar

Solution: Unit of vr = m/sm = m2 /s 

                 Unit of v2 = (m/s)2 = m2 /s2   

                 Unit of v/t a = (m/s) /s= m /s2   

                 Unit of at = (m/s2) s= m /s  

                 Unit of ½ a(t)2 = (m/s2) s2 = m  

                 Unit of vt = (m/s) s= m   (answer) 

                 Unit of 2ar = (m/s2) m= m2 /s2 

Part (c) Are the units of the formula v2 = 2ar dimensionally consistent? Select the single best answer. MultipleChoice : 1) no – v2 has units of m/s; 2ar has units of 1/s2 2) no – v2 has units of m2/s2; 2ar has units of 1/s2 3) no – v2 has units of m2/s2; 2ar has units of m/s2 4) no – v2 has units of m/s; 2ar has units of m/s2

5) yes – both v2 and 2ar have units of m/s 6) yes – both v2 and 2ar have units of m2/s2 7) no – v2 has units of m/s; 2ar has units of m2/s2

Solution: Unit of LHS is the unit of v2 = (m/s)2 = m2/s2 

                 Unit of RHS is the unit of 2ar = (m/s2)m = m2/s2 

                 So the formula is dimensionally consistent and have units of m2/s2 

 

Part (d) Are the units of the formula ma = mv2/2 dimensionally consistent? Select the single best answer. MultipleChoice : 1) yes – both ma and 1/2mv2 have units of kg•m2/s2 2) no – ma has units of kg•m/s and 1/2mv2 has units of kg•m2/s2 3) no – ma has units of kg•m2/s2 and 1/2mv2 has units of kg•m/s2 4) yes – the units of m cancel on both sides. 5) yes – both ma and 1/2mv2 have units of kg•m/s 6) no – ma has units of kg•m/s2 and 1/2mv2 has units of kg•m2/s2 7) yes – both ma and 1/2mv2 have units of kg•m/s2

Solution: Unit of LHS is the unit of ma = Kg (m/s2 )= Kg m/s2 

                 Unit of RHS is the unit of mv2/2 = Kg(m/s)2 = Kgm2/s2 

                 So the formula is dimensionally inconsistent 

 

Part (e) Which of the following will have units of kg•m2/s2? Select all that apply. MultipleSelect : 1) ma∆t 2) ma 3) mar 4) mv 5) 1/2mv2 6) ma/∆t

 

Solution:  Unit of mat = Kgm/s2s = Kgm /s 

                  Unit of ma = Kgm/s2 = Kgm /s2 

                  Unit of mar = Kgm/s2m = Kgm2 /s2   (yes) 

                  Unit of mv = Kgm/s = Kgm /s   

                  Unit of ½ mv2= Kg(m/s)2 = Kgm2 /s2   (yes) 

                  Unit of ma/t = Kg m/s2)/s = Kgm /s3 

 

 

Part (f) Given the formula p = mv, what are the units of p?

Solution: Unit of p = Kg (m/s) = Kgm/s

Part (g) Given the formula F = ma, what are the units of F?

Solution: Unit of F = Kg (m/s2) = Kgm/s2

Part (h) Given the formula E = 1/2mv2, what are the units of E? 

Solution: Unit of E = Kg (m/s)2 = Kgm2/s2

 

 

                  

   

Problem 4: Answer the following questions using your knowledge of calculus with the following table of base quantities, derived quantities and units. Use dimensional analysis to guide your thinking. 

 

 

  

 

Part (a) Determine the units of the differential dt. Select the single best answer. MultipleChoice : 1) s 2) m•s 3) No units. 4) ds

 

Solution: dt is just the difference in time, so the unit for dt should also be t. 

Part (b) Determine the units of the quantity dr/dt. Select the single best answer. MultipleChoice : 1) m/s 2) m 3) No units. 4) dm/ds 5) m2/m•s

 

Solution: Unit of dr is m and unit of dt is s , so the unit for dr/dt should m/s. 

Part (c) Determine the units of the quantity d2r/dt2. Select the single best answer. MultipleChoice : 1) m/s2 2) m3/(m•s2) 3) No units. 4) m3/(m2•s2) 5) m2/s2

 

Solution: d2r/dt2 =d (dr/dt) /dt .  d (dr/dt) is just the difference in dr/dt which should have the same unit 

as dr/dt.  So the unit of d(dr/dt) is m/s.  Unit of dt is s.  Unit of d2r/dt2 =d (dr/dt) /dt is (m/s)/s = m/s2. 

 

Part (d) Are the units of the formula a = dv/dx, where v is velocity and a acceleration, dimensionally consistent? Select the single best answer. MultipleChoice : 1) no – a has units of m/s2: dv(x)/dx has no units. 2) no – a has units of m/s2: dv(x))/dx has units of 1/s 3) yes – both a and dv(x)/dx have units of m/s. 4) yes – both a and dv(x)/dx have units of 1/s.

5) no – a has units of m/s2: dv(x)/dx has units of s. 6) yes – both a and dv(x)/dx have units of m/s2.

Solution: Unit of a is m/s2 and unit of dv/dx is (unit of v)/(unit of x) = (m/s)/m = 1/s, so the formula is 

dimensionally inconsistent. 

 

Part (e) Are the units of the formula for acceleration a = dv / dt dimensionally consistent? Select the single best answer. MultipleChoice : 1) yes – both a and dv(t)/dt have units of m/s2. 2) yes – both a and dv(t)/dt have units of m/s. 3) no – a has units of m/s2 and dv(t)/dt has units of 1/s. 4) no – a has units of m/s2 and dv(t)/dt has units of m. 5) no – a has units of m/s2: dv(t)/dt has units of m/s.

 

Solution: Unit of a is m/s2 and unit of dv/dt is (unit of v)/(unit of t) = (m/s)/s = m/s2, so the formula is 

dimensionally consistent and both sides have units m/s2. 

Part (f) Determine the units of the quantity ∫ v(t)dt. Select the single best answer. MultipleChoice : 1) m 2) m/s2 3) No units. 4) m2/s 5) m/s

  

Solution:  The integration is just summation, so the unit is just the unit of the terms to be summed.  Unit 

for vdt is (m/s)s = m, so the unit of ∫ v(t)dt should be m. (Note: v(t) means v is a function of t, not v times t)

Part (g) Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, and a is acceleration.

 

Solution: 

kgm sm/s)(kg v(t)dt 2

m ofUnit

/skgm sm/s)(kg dt [v(t)]2

m ofUnit

kgms smskg xdt m ofUnit

/skgm s

/smkg

s

(m/s)kg

dt

d[v(x)]

2

m ofUnit

(yes) m/s Kg s

m/skg

dt

dvm ofUnit

m/s Kg s

m/skg

dt

da(t)m ofUnit

kgm/s sm/skg adt m ofUnit

m/s Kg s

mkg

dt

dxm ofUnit

(yes) kgm/s m

/smkg

m

(m/s)kg

dx

d[v(x)]

2

m ofUnit

222

322222

2

32

2

22222

 

 

 

 

 

   

Problem 5: The units on each side of an equation should be the same, or the equation is wrong. You will determine which units belong on each side of an equation.

 

 

 

 

 

 

 

 

Part (a) What are the units on both sides of the equation: v2 = 2ax? MultipleChoice : 1) m•s/kg 2) m/s 3) kg•m/s2. 4) m3/s2. 5) m2/s2 6) kg2/m3. Solution: Unit of LHS = (m/s)2 = m2/s2 Unit of RHS = (m/s2)m = m2/s2 Unit of LHS = Unit of RHS Part (b) What units are on both sides of the equation: x = vt + 1/2at2? MultipleChoice : 1) s•m 2) m/s 3) kg•m 4) s2/m 5) m 6) m2/s Solution: Unit of LHS = m For RHS, unit of vt = (m/s)s = m unit of ½ at2 = (m/s2)s2 = m All have the same unit. Part (c) What are the units on both sides of the equation mv2 = mgh? MultipleChoice : 1) m•kg2 2) kg•m/s 3) s•kg/m

4) m/s 5) kg•m2/s2 6) s2/m2 Solution: Unit of LHS = kg(m/s)2 = kgm2/s2 Unit of RHS = kg (m/s2)m = kgm2/s2 Unit of LHS = Unit of RHS Part (d) What are the units on each side of the equation GMm/x2 = mv2/x MultipleChoice : 1) m•s2/kg 2) kg•m/s 3) s•m/kg 4) kg•m2/s 5) kg•m/s2 6) kg•m2/s3 Solution: Unit of LHS = m3/kgs2kgkg/m2 = mkg/s2 Unit of RHS = kg(m/s)2/m = kgm/s2 Unit of LHS = Unit of RHS

 

   

Problem 6: Answer the following questions about dimensional analysis.

Part (a) Which of the following are dimensionally consistent? There may be more than one correct answer.

 

Unit of LHS = 1/s Unit of RHS = (m/s2)/m = 1/s2 Unit of LHS Unit of RHS Inconsistent 

Unit of LHS = m Unit of RHS = (m/s)/(m/s2) = s Unit of LHS Unit of RHS Inconsistent 

 

Unit of LHS = (m/s)2 = m2/s2 Unit of RHS = (m/s2)m= m2/s2 Unit of LHS = Unit of RHS Consistent 

 

 

Unit of LHS = s2 Unit of RHS = (m/s2)/m= 1/s2 Unit of LHS Unit of RHS Inconsistent 

 

 

Unit of LHS = s Unit of RHS = [m/ (m/s2)]1/2 =[ s2]1/2 = s Unit of LHS = Unit of RHS Consistent 

 

 

Unit of LHS = s Unit of RHS = (m/s2) / (m/s) =1/ s Unit of LHS Unit of RHS Inconsistent 

   

Problem 7: A car is traveling at a speed of 25 m/s. 

Part (a) What is its speed in kilometers per hour?

Solution:

km/hr 90

km/hr 103600 25

m/hr 3600 25 m/s 253-

 

Part (b) Suppose the speed limit is 85 km/h. How many kilometers per hour would the car need to slow down to be going the speed limit?

Solution: 

The car has to slow down by 90 km/hr – 85 km/hr = 5km/hr. 

   

Problem 8: A particle’s position along the x-axis is described by the function x(t) = A t + B t2,

where t is in seconds, x is in meters, and the constants A and B are given below.

 

Part (a) Input an expression for the velocity of the particle as a function of time.

Solution:

2Bt A ) tB +A t (dt

d

dt

dx v

tB +A t = x(t)

2

2

 

Part (b) At what time, in seconds, is the particle’s velocity zero?

.2B

A- when tzero is velocity sparticle' The

2B

A- t

0 2Bt A 0v

2Bt A v

   

 

Problem 9: A particle’s position along the x-axis is described by x(t) = A t + B t2,

where t is in seconds, x is in meters, and the constants A and B are given below.

Part (a) What is the velocity, in meters per second, of the particle at t1 = 3.0 s?

Solution: 

 

6BA 3))((2B A vs,0.3At t

2Bt A ) tB +A t (dt

d

dt

dx v

tB +A t = x(t)

1

2

2

 

Part (b) What is the velocity, in meters per second, of the particle when it is at the origin (x = 0) at t0 > 0?

Solution: 

We need to figure out at what time the particle is at the origin first. 

A

2BA )

A

B-)((2B A v time,At this

2Bt A dt

dx v

negative) isA and positive is Bgiven that the(note A

B- t 0, trequires problem The

A

B-or t 0 t

0Bt +A or 0 t

0 tBt) +(A

0 tB +A t 0x

tB +A t = x(t)

2

2

2

 

 

   

Problem 10: A particle’s position along the x-axis is described by

x(t) = A t + B t2, where t is in seconds, x is in meters, and the constants A and B are given below. Part (a) What is the position, in meters, of the particle when the velocity is zero?

Solution:

4B

A-

4B

A +

2B

A-

)4B

A( B +

2B

A-

)2B

A(- B + )

2B

AA(- tB +A t = x time,At this

.2B

A- when tzero is velocity sparticle' The

2B

A- t

0 2Bt A 0v

2Bt A ) tB +A t (dt

d

dt

dx v

tB +A t = x(t)

2

22

2

22

22

2

2

   

Problem 11: Find the following for path D in the figure:

 

 

 

 

 

 

 

 

 

Part (a) The distance traveled in meters. Solution: Distance traveled = 6 + 2 = 8m Part (b) The magnitude of the displacement from start to finish in meters.

        Solution: |x| = |5m – 9m| = |-4m| = 4m Part (c) The displacement from start to finish in meters.       Solution: x = 5m – 9m = -4m

Problem 12: Position vs time is shown in the figure.

 

(a) Take the slope of the curve in the figure to determine the velocity at t = 10.0 s, in meters per second.

Since 10s is between 9.7s and 19.8s, so we will take the velocity at t=10s as the average velocity 

between 9.7s and 19.8s: 

m/s 211.01.10

13.2

7.98.19

4.77-6.9 Velocity

 

 

Problem 13: Please answer the following questions about displacement vs. time graphs.

Part (a) Which of the following graphs represents an impossible motion?

The following is not possible as a particle cannot be at more than one position at the same time:

Cannot have three displacements at 

this time!

Part (b) Which graph has only negative velocity?

 

 

 

 

 

 

The following is the only given graph with a negative slope all the time, so it has only negative velocity.

 

 

 

 

 

Part (c) Which graph represents an object being stationary for periods of time? 

 

 

 

 

 

 

 

 

 

 

 

 

The following graph represents an object being stationary for periods of time:

 

 

 

 

 

 

 

 

 

   

Stationary for periods of time (not just instantaneously) 

Problem 14: Consider the position at various times of an object, given in the table.

 

 

 

 

 

 

 

 

Part (a) Select the best position graph representing the information given in the table. 

 

   

 

   

 

 

Note that time is increased regularly by 2.5s for each pair of data, but the increment in position is 

becoming greater and greater, so it has to be represented by the following graph: