UNIT 1 Motion Graphs LyzinskiPhysics x t Days 3 - 4

Day #3 * instantaneous velocity * d-t vs. x-t graphs (what’s the difference???)

Definition Instantaneous Velocity (v) – the velocity of an object at a precise moment in time. v = lim(  x/  t) t0t0 Honors Only!!!

Just what is meant by “instantaneous” velocity? tt tt tt tt tt To find the average velocity between two points in time, we find the slope of the line connecting these two points, thus finding the change in position (rise) over the change in time (run). As the two points move closer together, we find the average velocity for a smaller time interval. As the two points move EVEN CLOSER together, we find the average velocity for an EVEN SMALLER time interval. Finally, “in the limit” that the time interval is infinitely small (or approximately zero), we find the velocity at a single moment in time.  Hence the term “instantaneous velocity”

To find instantaneous velocities, we still use the concept of slope. But we find the slope OF THE TANGENT TO THE CURVE at the time in question Definition Tangent to a Curve – a line that intersects a given curve at exactly one point.

Good Tangents Bad Tangents 

How to find the instantaneous velocity of a specific time interval from an x-t graph … x(m) t (s) Draw the tangent to the curve at the point in question. Then, find the slope of the tangent. Slope = rise/run = 15 m / 9 s = 1.7 m/s (approx) Example: What is the instantaneous velocity at t = 20 seconds? YOU MUST SPECIFY WHICH POINTS YOU USED WHEN FINDING THE SLOPE!!!! (24, 30) (15, 15)

How to find the instantaneous velocity of a specific time interval from an x-t graph … x(m) t (s) Example: What is the instantaneous velocity at t = 5? If the pt lies on a segment, find the slope of the segment. Slope = 5 m / 10 s = 0.5 m/s (0,5) (10,10) YOU MUST SPECIFY WHICH POINTS YOU USED WHEN FINDING THE SLOPE!!!!

How to find the instantaneous velocity of a specific time interval from an x-t graph … x(m) t (s) Draw the tangent to the curve at the point in question. Then, find the slope of the tangent. Slope = 0 (object at rest) Example: What is the instantaneous velocity at t = 25 seconds?

x-t graphs t (sec) x (m) t 1 t 2 t 3 x2x1x3x2x1x Slope of line segment

Tangent to the curve has a slope of -26m / 13.5s = m/s THEREFORE, v = m/s and s = 1.93 m/s (approximately) Open to in your Unit 1 packet 1 Tangent to the curve has a slope of +22m / 22sec = 1 m/s (13.5,-20) (0,6) (33,2) (11,-20)

HONORS ONLY Given a function for position, like x(t) = 3t 2 + 3t – 6, find the instantaneous velocity at a give time (like t = 2 sec) using your graphing calculator. Step #1) type the equation into “y =“ as y = 3x 2 + 3x - 6 Step #2) Use “2 nd -Trace-6” to use the “dy/dx” function Step #3) Since “dy/dx” is really dx/dt, type in “2” and hit enter to get the instantaneous velocity at t = 2 sec. ANSWER: dx/dt = 15 m/s.

Day #4 * Average Acceleration

Definition Average Acceleration ( a ) – the change in an object’s velocity divided by the elapsed time……..IN OTHER WORDS, the rate of change of an object’s velocity. a =  v/t

Find the acceleration of each object 1)An object is moving at 40 m/s when it slows down to 20 m/s over a 10 second interval. 2)An object is moving at -40 m/s, and 5 seconds earlier it was moving at -10 m/s. 3)An object travelling at -10 in/min is moving at +20 in/min 2 minutes later. 4) An object moving at -30 mph is moving at -20 mph 10 hours later. a =  v/  t = (20 – 40m/s) / 10sec = -2 m/s 2 a =  v/  t = [-40 – (-10m/s)] / 5sec = -6 m/s 2 a =  v/  t = [20 – (-10 in/min)] / 2min = +15 in/min 2 Slows down Negative accel Speeds up Negative accel Speeds up + Accel a =  v/  t = [-20 – (-30 mi/h)] / 10hr = + 1mi/h 2 Slows down + Accel

So, how can something slow down and have a positive acceleration? An object moving at -30 mph is moving at -20 mph 10 hours later. Its speed (30mph vs. 20 mph) clearly decreases. * remember, speed is |v| As time marches on, the velocity become MORE positive (b/c -20mph is more positive than -30mph)  THEREFORE,  v is +

What do the “units” of acceleration mean? m/s 2 3 m/s 2 means that your velocity increased by 3 m/s every second km/min 2 means that your velocity decreased by 0.1 km/min every minute that you were moving. m/s/s m/s per second t (sec)v (m/s) t (min)v (km/min)

The Key Equations Displacement: Velocity: Acceleration: AVERAGE INSTANTANEOUS Really just tangents to the curve at a point.

Open to in your Unit 1 packet 2 -1 m/s1 m/s +4 m/s4 m/s +9 m/s9 m/s +14 m/s14 m/s +19 m/s19 m/s

Call right + and left – v 1 = 5 m/s right = + 5 m/sv 2 = 4.8 m/s left = -4.8 m/s a = (v 2 – v 1 ) / t = (-4.8 – 5) /.002sec = -4,900 m/s 2 = 4,900 m/s 2 left v 1 = 60 mi/hv 2 = 0 mi/ha = -5 mi/h 2 a = (v 2 – v 1 ) / t  -5 = (0 – 60) / t  t = 12 s