Lecture 5: Vectors & Motion in 2 Dimensions

Lecture 5: Vectors &Motion in 2 Dimensions

Questions of Yesterday 2) I drop ball A and it hits the ground at t1. I throw ball B horizontally (v0y = 0) and it hits the ground at t2. Which is correct? a) t1 < t2 b)t1 > t2 c) t1 = t2

Questions of Yesterday 1) Can a vector A have a component greater than its magnitude A? YES b) NO 2) What are the signs of the x- and y-components of A + B in this figure? a) (x,y) = (+,+) b) (+,-) c) (-,+) d) (-,-) A B

Dr = rf - ri 2 Dimensions Displacement in 2 Dimensions Position vectors no longer accounted for by + and - Displacement = change in position vector of object = Dr Dx = xf - xi 1 Dimension y (m) Dr ri object path rf x (m)

Velocity in 2 Dimensions Average Velocity Dr Dt rf - ri tf - ti vav = = y (m) ti Instantaneous Velocity Dr ri tf Dr Dt lim Dt -> 0 v = object path rf x (m)

Dv Dt lim Dt -> 0 a = Acceleration in 2 Dimensions Average Acceleration Dv Dt vf - vi tf - ti = = aav vy (m/s) ti Instantaneous Acceleration Dv vi tf object’s instant. velocity vf vx (m/s)

Acceleration in 2 Dimensions Average Acceleration Dv Dt vf - vi tf - ti = = aav If a car is going North at a constant speed and makes a left turn while maintaining its constant speed and then continues West at the same speed… does the car accelerate during this trip? A runner is running on a circular track at constant speed? Is she accelerating?

Acceleration in 2 Dimensions Average Acceleration Dv Dt vf - vi tf - ti = = aav Velocity is a vector with both magnitude & direction, so… An object can accelerate by either changing its SPEED or changing its DIRECTION

Projectile Motion Motion in 2 Dimensions under constant gravitational acceleration Horizontal component of velocity is constant over entire path! vx = v0x No acceleration in horizontal direction

Projectile Motion Motion in 2 Dimensions under constant gravitational acceleration Vertical component of velocity constantly changing due to gravitational acceleration in -y direction v0y --> 0 -> -v0y

t= 1 s t= 2 s t= 3 s t= 4 s t= 5 s Projectile Motion Initial velocity in horizontal direction no gravity Initial velocity in horizontal direction with gravity Initial velocity in vertical direction with gravity Horizontal and Vertical motions are completely independent of each other!! Motion in one direction has NO EFFECT on motion in the other direction

t= 1 s t= 2 s t= 3 s t= 4 s t= 5 s Important Features of Projectile Motion Acceleration is ALWAYS -9.80 m/s2 in the vertical direction Parabolic motion is symmetric vf = -v0 At the top of the trajectory: t = 1/2 of total time x = 1/2 of total horizontal range Total time of trajectory is independent of horizontal motion

t= 1 s t= 2 s t= 3 s t= 4 s t= 5 s Important Features of Projectile Motion At what point in the object’s trajectory is the speed a minimum? What about velocity?

q 2D Motion under Constant Acceleration Because x and y motions are independent… we can apply 1D equations for constant acceleration motion separately to each x- and y- direction But…. v0 has both x- and y-components Need to separate v0 into x- and y- components v0x = v0cosq v0y = v0sinq

Horizontal Motion Vertical Motion vx = v0x + axt Dx = v0xt + 1/2axt2 vx2 = v0x2 + 2axDx vy = v0y + ayt Dy = v0yt + 1/2ayt2 vy2 = v0y2 + 2ayDy 2D Motion under Constant Acceleration Recall equations for 1D motion under constant acceleration v = v0 + at Dx = v0t + 1/2at2 v2 = v02 + 2aDx 2D motion equivalent to superposition of two independent motions in the x- and y-directions

v0x = v0cosq ax = 0 vx = v0cosq = constant Dx = v0xt = (v0cosq)t t= 1 s t= 2 s t= 3 s Time is still determined by y-direction motion! t= 4 s t= 5 s Horizontal Motion of Projectile vx = v0x + axt Dx = v0xt + 1/2axt2 vx2 = v0x2 + 2axDx

v0y = v0sinq ay = g = -9.80 m/s2 vy = v0sinq + gt Dy = (v0sinq)t + 1/2gt2 vy2 = (v0sinq)2 + 2gDy q Pay attention to sign convention! Vertical Motion of Projectile vy = v0y + ayt Dy = v0yt + 1/2ayt2 vy2 = v0y2 + 2ayDy

Equations for Motion of Projectile vy = v0sinq + gt Dy = (v0sinq)t + 1/2gt2 vy2 = (v0sinq)2 + 2gDy vx = v0cosq = constant Dx = v0xt = (v0cosq)t Vertical Component Motion Horizontal Component Motion v = (vx2 + vy2)1/2 tanq = vy/vx q = tan-1(vy/vx) -90 < q < 90 Combined 2D Motion

t= 1 s t= 2 s t= 3 s t= 4 s t= 5 s Problem #1 A projectile falls beneath the straight-line path it would follow if there were no gravity. How many meters does it fall below this line if it has been traveling for 1 s? For 2 s? Does your answer depend on the angle at which the projectile is launched? What about the speed?

Questions of the Day • 2) Two projectiles are thrown with the same initial speed, one at an angle q with respect to the ground and the other at an angle 90o - q. Both projectiles strike the ground at the same distance from the projection point. Are both projectiles in the air for the same length of time? • a) YES • b) NO • 1) A heavy crate is dropped from a high-flying airplane as it flies directly over your shiny new car? Will your car get totaled? • YES b) NO

Lecture 5: Vectors & Motion in 2 Dimensions