Speed or Velocity Time Graphs

Do In Notes: Sketch a d – t graph for object moving at constant speed Do In Notes: Sketch a d – t graph for object moving at constant speed. Now sketch a speed time graph showing the same motion. d v t t

Constant Velocity/Speed

Sketch a distance – t graph for object starting from rest and speeding up with constant acceleration. Now sketch a speed time graph showing the same motion. v d t t

Average Velocity is the midpoint between 2 speeds. vf + vi 2 Constant / Uniform Acceleration. Slope = Constant Acceleration of straight line on v-t graph. Average Velocity is the midpoint between 2 speeds. vf + vi 2

What is the average speed between 1 – 3 seconds? 1. What is the acceleration of this object? Slope = accl. Dy/ Dx (50 – 10) m/s (5 – 1) s What is the sign of accl? What is the average speed between 1 – 3 seconds? What is the average speed between 3 – 5 seconds?

2. What is the acceleration between: 0 – 3 seconds, 5-10 seconds? Slope = 2 m/s2. 0.

3. What’s going on here?

Velocity Time Graphs Vector Nature sign of acceleration

Sign of velocity.

Finding Distance or Displacement on V-T graphs Speed Time = distance Velocity Time = displacement

Displacement = Area Under Curve at specific time Displacement = Area Under Curve at specific time. Drop a vertical to the X axis. 4. What is the displacement at 20 sec? A = bh = (20s)(30m/s)

5. What is the displacement at 5 s? Area = bh = (5 s) x (1 m/s) = 5 m.

6. What is the displacement at 4 seconds? A = ½ bh = ½ (4s)(40 m/s) = 80 m.

7. What is the displacement at 10 s? A1 = 1/2bh = 1/2(4s)(8m/s) = 16 m A2 = bh = (6 s) (8 m/s) = 48 m A tot = 64 m.

Return to start point

8. How can you tell when object is back to starting point? Positive displacement = negative displacement. Tot displacement = 0.

9. At what time does the object return to the starting point? At 5 seconds d = bh =(5s)(1m/s)= + 5 m. From 5 – 10 seconds d = bh =(5s)(-1m/s)= - 5 m. At t = 10 s.

Given the v – t graph below, sketch the acceleration – t graph for the same motion.

Acceleration – time Graphs What is the physical behavior of the object? Slowing down pos direction, constant vel neg accel.

d-t: v-t: a-t: slope = velocity area ≠ . slope = accl area = displ area = D vel vf – vi.

Hwk. handout “Motion Graph Prac”.

Objects Falling Under Gravity

Freefall Gravity accelerates uniformly masses as they fall and rise. Earth’s acceleration rate is 9.81 m/s2 – very close to 10 m/s2.

Falling objects accelerate at the same rate in absence of air resistance

But with air resistance

Fortunately there is a “terminal fall velocity.” After a while, the diver falls with constant velocity due to air resistance. Unfortunately terminal fall velocity is too large to live through the drop.

Apparent Weightlessness Objects in Free-fall Feel Weightless

What is the graph of a ball dropped?

What do the d-t, v-t, and a – t, graphs of a ball thrown into the air look like if it is caught at the same height?

A ball is thrown upward from the ground level returns to same height. a is -9.81, the ball is accelerating at constant 9.81 m/s2. d = ball’s height above the ground velocity is + when the ball is moving upward Why is acceleration negative? Is there ever deceleration?

Free-fall Assumptions Trip only in the air. Trip ends before ball caught. -Symmetrical Trip time up = time down -Top of arc: v = 0, a = ?? -On Earth g = -9.81 m/s2. Other planets g is different.

Solving: Use accl equations replace a with -g. List given quantities & unknown quantity. Choose accl equation that includes known & 1 unknown quantity. Be consistent with units & signs. Check that the answer seems reasonable Remain calm

Practice Problem. 1. A ball is tossed upward into the air from the edge of a cliff with a velocity of 25 m/s. It stays airborne for 5 seconds. What is its total displacement?

vi = +25 m/s a = g = -9.81 m/s2. t = 5 s. d = ? d= vit + ½ at2. (25m/s)(5s) + 1/2(-9.81 m/s2)(5 s)2. 125 m - 122 . 6 = +2.4m. It is 2.4m above the start point.

It will be below the start point. 2. If the air time from the previous problem is increased to 5.2 seconds, what will be the displacement? d = vit + ½ at2. -2.6 m It will be below the start point.

Ex 3. A 10-kg rock is dropped from a 7- m cliff Ex 3. A 10-kg rock is dropped from a 7- m cliff. What is its velocity just before hitting the ground? d = 7m a = -9.81 m/s2. vf = ? Hmmm vi = 0. vf 2 = vi2 + 2ad vf 2 = 2(-9.81m/s2)(7 m) vf = -11.7 m/s (down)

4. A ball is thrown straight up into the air with a velocity of 25 m/s   4. A ball is thrown straight up into the air with a velocity of 25 m/s. Create a table showing the balls position, velocity, and acceleration for each second for the first 5 seconds of its motion.  T(s) d v (m/s) a (m/s2) 0 0 25 -9.81 1 20 15.2 -9.81 2 30 5.4 -9.81 3 31 -4.43 -9.81 5 2.4 -24 -9.91

Read Text pg 60-64 Do prb’s pg 64# 1-5 show work.

Mech Universe: The Law of Falling Bodies: http://www.learner.org/resources/series42.html?pop=yes&pid=549#