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Rhoda Wilkinson
9 years ago
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1 Problem (Q1): Evaluate each of the following to three significant figures and express each answer in SI units: (a) (0.631 Mm)/(8.60 kg) 2 (b) (35 mm) 2 *(48 kg) 3 (a) Mm / 8.60 kg m 8532 m = = kg kg = m / kg = 8.53 km / kg (b) 35 mm 48 kg = m 48 kg = 135 m / kg

60 kg) 2 (b) (35 mm) 2 *(48 kg) 3 (a) 0.631 Mm / 8.60 kg 2 6 0.

2 A sphere is fired downwards into a medium with an initial speed of 27 m>s. If it experiences a deceleration of a = (-6t) m>s 2, where t is in seconds, determine the distance traveled before it stops. Velocity: v 0 = 27 m>s at t 0 = 0 s. Applying Eq. 12 2, we have A+TB dv =adt v t dv = -6tdt L27 v = A27-3t 2 B m>s (1) At v = 0, from Eq. (1) Distance Traveled: s at. Using the result v = 27-3t 2 0 = 0 m t 0 = 0 s and applying Eq. 12 1, we have A+TB At t = 3.00 s, from Eq. (2) 0 = 27-3t 2 t = 3.00 s ds =vdt s t ds = A27-3t 2 Bdt L0 s = A27t -t 3 B m s = 27(3.00) = 54.0 m (2) This work is protected by United States copyright laws

Applying Eq. 12 2, we have A+TB dv =adt v t dv = -6tdt L27 v = A27-3t 2 B m>s (1) At v = 0, from Eq. (1) Distance Traveled: s at.

3 A particle is moving along a straight line such that when it is at the origin it has a velocity of 4 m>s. If it begins to decelerate at the rate of a =1-1.5v 1>2 2 m>s 2, wherevis in m>s, determine the distance it travels before it stops. a = dv dt = -1.5v1 2 v t v dv = -1.5dt L4 2v 1 2 v 4 = -1.5tt 0 2av 1 2-2b = -1.5t v = (2-0.75t) 2 m>s (1) s t t ds = (2-0.75t) 2 dt = (4-3t t 2 )dt L0 laws s = 4t - 1.5t t 3 (2) copyright From Eq. (1), the particle will stop when States 0 = (2-0.75t) United 2 t = s s t=2.667 = 4(2.667) - 1.5(2.667) (2.667) 3 = 3.56by 3.56 m This work is protected

5v1 2 v t v - 1 2 dv = -1.5dt L4 2v 1 2 v 4 = -1.5tt 0 2av 1 2-2b = -1.5t v = (2-0.75t) 2 m>s (1) s t t ds = (2-0.75t) 2 dt = (4-3t + 0.

4 * A particle travels to the right along a straight line with a velocity v = [5>14 +s2] m>s, where s is in meters. Determine its deceleration when s = 2 m. v = v dv = a ds dv = s -5 ds (4 + s) 2 5 (4 + s) a - 5 ds (4 + s) 2b = a ds a = - 25 (4 + s) 3 When s = 2 m a = m>s 2 This work is protected by United States copyright laws

5 As a train accelerates uniformly it passes successive kilometer marks while traveling at velocities of 2 m>s and then 10 m>s. Determine the train s velocity when it passes the next kilometer mark and the time it takes to travel the 2-km distance. Kinematics: For the first kilometer of the journey, v 0 = 2 m>s, v = 10 m>s, s 0 = 0, and s = 1000 m. Thus, A: + B v 2 =v a c (s -s 0 ) 10 2 = a c (1000-0) a c = m>s 2 For the second kilometer, v 0 = 10 m>s, s 0 = 1000 m, s = 2000 m, and a m>s 2 c =. Thus, A: + B v 2 =v a c (s -s 0 ) v 2 = (0.048)( ) v = 14 m>s For the whole journey, v, v = 14 m>s, and m>s 2 0 = 2 m>s a c =. Thus, A: + B v =v 0 +a c t 14 = t t = 250 s work is protected by United States copyright laws and is provided solely for the use of instructors Ain teaching

Kinematics: For the first kilometer of the journey, v 0 = 2 m>s, v = 10 m>s, s 0 = 0, and s = 1000 m. Thus, A: + B v 2 =v 2 0 + 2a c (s -s 0 ) 10 2 = 2 2 + 2a c (1000-0) a c = 0.

Scalar versus Vector Quantities. Speed. Speed: Example Two. Scalar Quantities. Average Speed = distance (in meters) time (in seconds) v =

Scalar versus Vector Quantities Scalar Quantities Magnitude (size) 55 mph Speed Average Speed = distance (in meters) time (in seconds) Vector Quantities Magnitude (size) Direction 55 mph, North v = Dx