Motion in One Dimension (Velocity vs. Time) Chapter 5.2

What is instantaneous velocity?

What effect does an increase in velocity have on displacement?

Instead of position vs. time, consider velocity vs. time. Relatively

How can be determined from a v vs. t graph? Measure the under the curve. d = v*t Where t is the x component v is the y component Time Velocity A2 A1

Measuring displacement from a velocity vs. time graph.

What information does the slope of the velocity vs. time curve provide? sloped curve = velocity (Speeding up). sloped curve = velocity (Slowing down). sloped curve = velocity. Time Velocity Acceleration C Time Velocity Acceleration A Time Velocity Acceleration B

What is the significance of the slope of the velocity vs. time curve? Since velocity is on the y-axis and time is on the x-axis, it follows that the slope of the line would be: Therefore, slope must equal . Time Velocity

Acceleration determined from the slope of the curve. What is the acceleration from t = 0 to t = 1.7 seconds? m = Since m = : a = ___/___ Slope =

Determining velocity from acceleration If acceleration is considered constant: a = __/__ = (__ – __)/(__ – __) Since ti is normally set to 0, this term can be eliminated. Rearranging terms to solve for vf results in: __ = __ + a__ Time Velocity Positive Acceleration Velocity

Position, velocity and acceleration when t is unknown. __ = __ + ½ (vf + vi)*t (1) vf = vi + at (2) Solve (2) for t: t = (__ – __)/__ and substitute back into (1) df = di + ½ (vf + vi)(__ – __)/__ By rearranging: __ = __ + 2__*(_____) (3)

Substitute (2) into (1) for vf df = di + ½ (__ + __ __ + __)*t Alternatively, If time and acceleration are known, but the final velocity is not: df = di + ½ (vf + vi)*t (1) vf = vi + at (2) Substitute (2) into (1) for vf df = di + ½ (__ + __ __ + __)*t df = di + __ __ + __ __ __ (4)

Formulas for Motion of Objects Equations to use when an accelerating object has an initial velocity. Form to use when accelerating object starts from rest (vi = 0). d = ½ (vi + vf) t d = ½ vf t vf = vi + at vf = at d = vi t + ½ a(t)2 d = ½ a(t)2 vf2 = vi2 + 2ad vf2 = 2ad

Acceleration due to All falling bodies accelerate at the rate when the effects of due to can be ignored. Acceleration due to is caused by the influences of Earth’s on objects. The acceleration due to is given the special symbol . The acceleration due to is a to the surface of the earth. = __/__

Example 1: Calculating Distance A stone is dropped from the top of a tall building. After 3.00 seconds of free-fall, what is the displacement, y of the stone? Data y ? a = g -9.81 m/s2 vf n/a vi 0 m/s t 3.00 s

Example 1: Calculating Distance From your reference table: d = ____ + ____ Since vi = __ we will substitute __ for __ and __ for __ to get: __ = ____

Example 2: Calculating Final Velocity What will the final velocity of the stone be? Data y -44.1 m a = g -9.81 m/s2 vf ? vi 0 m/s t 3.00 s

Example 2: Calculating Final Velocity Using your reference table: vf = __ + __ __ Again, since vi = __ and substituting __ for __, we get: vf = __ __ vf = Or, we can also solve the problem with: vf2 = ___ + _____, where vi = __

Example 3: Determining the Maximum Height How high will the coin go? Data y ? a = g -9.81 m/s2 vf 0 m/s vi 5.00 m/s t

Example 3: Determining the Maximum Height Since we know the initial and final velocity as well as the rate of acceleration we can use: ___ = ___ + ______ Since Δ__ = Δ__ we can algebraically rearrange the terms to solve for Δ__.

Example 4: Determining the Total Time in the Air How long will the coin be in the air? Data y 1.28 m a = g -9.81 m/s2 vf 0 m/s vi 5.00 m/s t ?

Example 4: Determining the Total Time in the Air Since we know the and velocity as well as the rate of we can use: __ = __ + __ __, where __ = __ Solving for t gives us: Since the coin travels both and , this value must be to get a total time of s

Key Ideas Slope of a velocity vs. time graphs provides an objects . The area under the curve of a velocity vs. time graph provides the objects . Acceleration due to gravity is the for all objects when the effects of due to wind, water, etc can be ignored.

Important equations to know for uniform acceleration. df = di + ½ (vi + vf)*t df = di + vit + ½ at2 vf2 = vi2 + 2a*(df – di) vf = vi +at a = Δv/Δt = (vf – vi)/(tf – ti)

Displacement when acceleration is constant. Displacement = area under the curve. Δd = vit + ½ (vf – vi)*t Simplifying: Δd = ½ (vf + vi)*t If the initial position, di, is not 0, then: df = di + ½ (vf + vi)*t By substituting vf = vi + at df = di + ½ (vi + at + vi)*t df = di + vit + ½ at2 vf d = ½ (vf-vi)t d = vit vi t