Speed or Velocity Time Graphs Do In Notes

Speed or Velocity Time Graphs

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

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

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/s 2. • 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. Drop a vertical to the X axis. 4. What is the displacement at 20 sec? A = bh = (20 s)(30 m/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 (4 s)(40 m/s) = 80 m.

7. What is the displacement at 10 s? • A 1 = 1/2 bh = 1/2(4 s)(8 m/s) = 16 m • A 2 = 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 =(5 s)(1 m/s)= + 5 m. • From 5 – 10 seconds d = bh =(5 s)(-1 m/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: – slope = velocity – area ≠. • v-t: – slope = accl – area = displ • a-t: – slope ≠. – 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/s 2 – very close to 10 m/s 2.

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. d = ball’s height above the ground velocity is + when the ball is moving upward Why is acceleration negative? Is there ever deceleration? a is -9. 81, the ball is accelerating at constant 9. 81 m/s 2.

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/s 2. 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/s 2. t = 5 s. d = ? • d= vit + ½ at 2. • (25 m/s)(5 s) + 1/2(-9. 81 m/s 2)(5 s)2. • 125 m - 122. 6 = +2. 4 m. • It is 2. 4 m above 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 at 2. • -2. 6 m • It will be below the start point.

Ex 3. A 10 -kg rock is dropped from a 7 - m cliff. What is its velocity just before hitting the ground? • d = 7 m • a = -9. 81 m/s 2. • vf = ? • Hmmm • vi = 0. • vf 2 = vi 2 + 2 ad • vf 2 = 2(-9. 81 m/s 2)(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. 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/s 2) • • • 0 20 30 31 2. 4 25 15. 2 5. 4 -4. 43 -24 -9. 81 -9. 91 0 1 2 3 5

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/series 42. html? pop=yes&pid=549#