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PHYSICS 149: Lecture 9 • Chapter 3 – 3.2 Velocity and Acceleration – 3.3 Newton’s Second Law of Motion – 3.4 Applying Newton’s Second Law – 3.5 Relative Velocity Lecture 9 Purdue University, Physics 149 1 Velocity (m/s) • The average velocity is the change in position (vector) divided by the change in time. Δr x f − x0 v= = Δt t f − t0 x(t) Δx Δt t • Instantaneous velocity is the limit of average velocity as ∆t gets small. It is the slope of the x(t) plot. G Δr G v = lim Δt →0 Δt Lecture 9 x(t) (t) t Purdue University, Physics 149 2 ILQ 1 • A go-kart travels around a circular track at a constant speed. Which of these is a true statement? (a) The go-kart has a constant velocity. (b) The go-kart has zero acceleration. A) Both (a) and (b) are true B) Neither (a) nor (b) is true C) (a) D) (b) Lecture 9 Purdue University, Physics 149 3 ILQ 2 If the average velocity of a car during a trip along a straight road is positive, is it possible for the instantaneous velocity att some time ti during d i th the ttrip i tto b be negative? ti ? A - Yes correct The car might have reversed for a little while along the trip creating a negative instantaneous velocity at the point. If the overall p displacement of the car is positive for that particular time interval than the average interval, velocity is positive as well. Lecture 9 B - No If the car is traveling in a straight g p path the velocity y will always be positive. The car needs to travel in the opposite direction to get a negative velocity. Purdue University, Physics 149 4 How to determine ∆x from v(t) • The total displacement ∆x during and time ∆t is the area under the graph v(t) • Assume 1D, and v constant: Δx vx = vav , x = Δt Δx = v1x Δt Lecture 9 v v1x Purdue University, Physics 149 Δx t1 t2 t 5 Acceleration (m/s2) • The average acceleration is the change in velocity divided by the change in time. v(t) Δv a= Δt Δv Δt t • Instantaneous acceleration is limit of average acceleration as ∆t gets small. It is the slope of the v(t) ()p plot. v(t) G Δv G a = lim Δt →0 Δt Lecture 9 Purdue University, Physics 149 t 6 Average Acceleration • Average acceleration is the change in velocity during g the time interval during g which the velocity y changed. " change" g in velocity = elapsed time • The x- and y-components of the average velocity are: • Typical units for the magnitude of acceleration are m/s2, km/s2, and so on. Lecture 9 Purdue University, Physics 149 7 Instantaneous Acceleration • Instantaneous acceleration is the average acceleration we measure when the time interval is infinitesimally short. • The x- and y-components of the instantaneous velocity are: Lecture 9 Purdue University, Physics 149 8 Direction of Acceleration • If speed is increasing, v and a are in same direction. direction • If speed is decreasing, v and a are in opposite direction. Lecture 9 Purdue University, Physics 149 9 ILQ: Acceleration If the velocity of some object is not zero, can its acceleration ever be zero? A) Yes B) No If the velocity is constant, the acceleration is zero ((slope p of v(t) ( ) is 0)). v(t) v(t) t Lecture 9 Purdue University, Physics 149 t 10 ILQ: Acceleration Is it possible for an object to have a positive velocity at the same time as it has a negative acceleration? A) Yes B) No Yes, when they have different signs, the object is slowing down. v(t) v(t) Δv Δt Lecture 9 t Purdue University, Physics 149 t 11 Position vs Time Plots • Gives location at any time • Displacement is change in position • Slope gives velocity x (m) 3 Position at t=3 t=3, x(3) = 1 4 Displacement between t=5 and t=1. Δx = -1.0 m 1.0 m - 2.0 m = -1.0 m t -3 Average g velocity y between t=5 and t=1. v = -0.25 m/s -1 m / 4 s = -0.25 m/s Lecture 9 Purdue University, Physics 149 12 Velocity vs Time Plots • Gives velocity at any time • Area gives displacement • Slope gives acceleration Velocity at t=2, v(2) = 3 m/s Displacement between t=0 t 0 and t=3: t 3: ∆x = 7.5 75m t=0 to t=1: ½ (3m/s) (1 s) = 1.5 m v (m/s) 3 1.5 6 4 t -33 t=1 to t=3: (3m/s) (2 s) = 6 m Average velocity between t=0 and t=3? v= 7.5 m / 3s = 2.5 m/s Change in v between t=5 and t=3. ∆v = -2 m/s – 3 m/s = -5 m/s A Average acceleration l ti b between t tt=5 5 and d tt=3: 3 a = -5 m/s / (2 s) = -2.5 m/s2 Lecture 9 Purdue University, Physics 149 13 Acceleration vs Time Plots • Gives acceleration at any time • Area gives change in velocity a (m/s2) 3 Acceleration at t=4 t=4, a(4) = -2 6 m/s2 24 Change in v between t=4 and t=1. ∆v = +4 m/s t=1-3: ∆v = (3m/s2)(2s) = 6 m/s t -3 t=3-4: ∆v = (-2m/s2)(1s) = -2 m/s Lecture 9 Purdue University, Physics 149 14 ILQ: Concepts • A skydiver is falling straight down, along the negative y direction. During the initial part of the fall, her speed increases from 16 to 28 m/s in 1.5 s. Which of the following is correct? A) v>0, a>0 v a B) v>0 v>0, a<0 C) v<0, a>0 D) v<0, a<0 y Í correct • During a later part of the fall, after the parachute has opened, her speed decreases from 48 to 26 m/s in 11 s. Which of the following is correct? A) B) C) D) Lecture 9 v>0, a>0 v>0, a<0 v<0, v<0 a>0 v<0, a<0 Í correct v a Purdue University, Physics 149 15 Dropped Ball A ball is dropped from a height of two meters above the ground. • Draw vy vs t 3 v A -2 3 4 t 3 B v 3 4 -2 4 x v C t 3 D -2 Lecture 9 v y v -2 t E t -2 Purdue University, Physics 149 4 4 t 16 Dropped Ball x A ball is dropped from a height of two ground. meters above the g t v • Draw v vs t • Draw x vs t • Draw a vs t t a t Lecture 9 Purdue University, Physics 149 17 Tossed Ball A ball is tossed from the ground up a height of two meters above the ground and falls back down. y • Draw v vs t 3 v -2 A 3 4 t 3 B 3 4 -2 v 4 v t E t Purdue University, Physics 149 4 -2 v x C t 3 D -2 Lecture 9 v -2 4 t 18 Tossed Ball A ball is tossed from the ground up a height of two meters above the ground and falls back down. x t v • Draw v vs t • Draw x vs t • Draw a vs t t a t Lecture 9 Purdue University, Physics 149 19 ILQ • Which car has a westward acceleration? A) B) C) D) E) a car traveling westward at constant speed a car traveling t li eastward t d and d speeding di up a car traveling westward and slowing down a car traveling li eastward d and d slowing l i d down a car starting from rest and moving toward the eastt Lecture 9 Purdue University, Physics 149 20 Newton’s Second Law of Motion • If there is a net force acting on an object, it experiences an acceleration in the direction of the net force, and the magnitude of the acceleration is the net force’s magnitude divided by the mass of the object. • In component form, • Units for force: 1 N = 1 kg⋅ g m/s2 Lecture 9 Purdue University, Physics 149 21 ILQ • A single force is acting on an object. Which type of motion is not physically allowed? A) B) C) D) the object speeds up the object is at rest the object slows down the object's motion changes direction Lecture 9 Purdue University, Physics 149 22 ILQ If an object is acted on by two finite constant forces, is it p possible for the object j to move at constant velocity? A) B)) C) D)) E) Lecture 9 No, it will accelerate Yes, the forces must be p perpendicular p No, it will follow a curved path Yes,, the forces must be equal q and opposite pp Yes, the forces must be in the same direction Purdue University, Physics 149 23 ILQ • A force f F acting i on a mass m1 results l iin an acceleration a1.The same force acting on a different mass m2 results in an acceleration a2 = 2a1. What is the mass m2? F A) • • • • m1 2m1 a1 B) m1 m2 F C) a2 = 2a1 1/2 m1 F=ma F= m1a1 = m2a2 = m2(2a1) Therefore, m2 = m1/2 Or in words…twice the acceleration means half the mass Lecture 9 Purdue University, Physics 149 24 Mass • Mass is a measure of an object’s inertia – the amount of resistance to changes in velocity. • Mass is an intrinsic property of an object, object so mass does not change wherever the object is (unlike weight). • The units of mass are g, kg, and so on • Objects with a larger amount of inertia (large mass) are harder to change the state of motion of compared to objects with a small amount of inertia (small mass) For the same net force, am (=|ΣF/m|) > aM (=|ΣF/M|) because m < M •m Lecture 9 M Purdue University, Physics 149 25 Newton’s Second Law F = ma A tractor T is pulling a trailer M with a constant acceleration. If the forward acceleration is 1.5 m/s2. Calculate the force on the trailer M (m=400Kg) due to the tractor T (m=500Kg). x – direction Fx = max T = max y T = (400 kg ) (1.5 m/s 2 ) N T T = 600 Newtons x W Lecture 9 Purdue University, Physics 149 26 Example A tractor t t T (m=500Kg) ( 500K ) is i pulling lli a ttrailer il M ((m=400Kg). 400K ) It starts from rest and pulls with constant force such that after 1 seconds its velocityy is 1m/s. Calculate the horizontal force y on the tractor due to the ground. x-direction: Tractor ΣF = ma Fw – T = mtractora Fw = T+ mtractora x-direction: di i T Trailer il ΣF = ma T = mtrailera Lecture 9 x N W T T N Fw W Combine: Fw = mtrailera + mtractora Fw = (m ( trailer+mtractor ) a Purdue University, Physics 149 27 Example A ttractor t T (m=500Kg) ( 500K ) is i pulling lli a ttrailer il M ((m=400Kg). 400K ) It starts from rest and pulls with constant force such that after 1 seconds its velocityy is 1m/s. Calculate the horizontal force y on the tractor due to the ground. x N Combine: Fw = mtrailera + mtractora Fw = (mtrailer+mtractor ) a a=Δv/Δt=(1m/s)/(1s)=1 W m/s2 T T N Fw W Constant C t t force f = constant a Fw=(900kg)×1 m/s2 =900N Lecture 9 Purdue University, Physics 149 28 ILQ • A wrench has a mass of 0.75 kg here on earth. It is shipped up to be used on the International Space Station. An astronaut on the Space Station will measure its mass to be: A)) B) C) D) 0 kg g 7.5 N 1.7 1 7 lbs 0.75 kg How about its weight? Lecture 9 Purdue University, Physics 149 29 Relative Velocity • W We often ft assume that th t our reference f frame f is i attached tt h d to t the th Earth. What happen when the reference frame is moving at a constant velocityy with respect p to the Earth? • The motion can be explained by including the relative velocity of the reference frame in the description of the motion. motion The ground speed of an airplane is the vector sum of the airspeed and the wind velocity. Using the air as the intermediate reference frame, ground speed is: Example airplanes G G G v PG = v PA + v AG Lecture 9 Purdue University, Physics 149 30 Relative Velocity • What is the relation between the velocity of an object as determined in one reference frame and the velocity of the same object as determined in another reference frame that is moving g with respect to the first reference frame? vAC = vAB + vBC where vAC is the velocityy of A relative to C,, vAB is the velocity of A relative to B, and vBC is the velocity of B relative to C. Lecture 9 Purdue University, Physics 149 31 Example where vWT is the velocity of Wanda relative to Tim, vTG is the velocity of Tim relative to Greg, and vWG is the velocity of Wanda relative to Greg. Lecture 9 Purdue University, Physics 149 32 ILQ • As a train travels with a velocity vTG = 15.5 m/s to right, Wanda walks toward the front of the train with a velocity vWT = 1.2 1 2 m/s to right. right What is her velocity relative to the ground? A)) B) C) D) 0 m/s 16.7 m/s, to right 14.3 m/s, to right impossible to tell Lecture 9 Purdue University, Physics 149 33 ILQ: Relative Velocity You are on a train traveling 40 mph North. If you walk 5 mph sideways across the car (W) (W), what is your speed relative to the ground? A) < 40 mph 5 40 Lecture 9 B) 40 mph C) >40 mph | v |= 40 + 5 = 40.3mph 2 Purdue University, Physics 149 2 34 ILQ: Time to Cross the River Three swimmers can swim equally fast relative to the water. Theyy have a race to see who can swim across a river in the least time. Relative to the water water, Beth (B) swims perpendicular to the flow, Ann (A) swims upstream, and C l (C) swims Carly i d downstream. t Which swimmer wins the race? A) Ann B) Beth C) Carly Lecture 9 A B C Beth uses all of her force in the 'y' direction, the direction of the other bank. Time to get across = width of river / y-component of velocity Purdue University, Physics 149 35