Chapter 2 Motion in One Dimension Ms. Hanan Anabusi

2-2 Acceleration Objectives: Describe motion in terms of changing velocity. Compare graphical representations of accelerated and nonaccelerated motions. Apply kinematic equations to calculate distance, time, or velocity under conditions of constant acceleration..

2-2 Acceleration Vocabulary: Acceleration Average acceleration Constant acceleration

Acceleration Acceleration describes the rate of change of velocity in a given time interval (how fast/slow an object speeds up or slows down) unit for acceleration: m/s2 m/s/s

Sample Problem B page 49 A shuttle bus slows down with an average acceleration of -1.8m/s2. How long does it take the bus to slow from 9.0 m/s to a complete stop. Solve Practice B #1-5 page 49

Acceleration Acceleration has both direction and magnitude. When the change in velocity is positive, acceleration is positive (vf > vi) [speeding up] When velocity is constant, acceleration is zero (vf = vi) When the change in velocity is negative, (vf < vi), acceleration is negative [slowing down] Acceleration can be described on a velocity-time graph.

Velocity-Time Graphs Constant positive acceleration (velocity is increasing), slope moves up and to the right. No acceleration (constant velocity), no slope: horizontal line Constant negative acceleration could mean the object is slowing down or moving in the opposite direction.

Uniform Acceleration Displacement of an object depends on acceleration, initial velocity, and time. For an object moving with constant acceleration, the average velocity is equal to the average of the initial velocity and the final velocity. Setting the two equations for average velocity equal to each other, and simplifying, we derive the formula for displacement with constant uniform acceleration: Dx = 1/2 (vi + vf) Dt displacement= 1/2(initial velocity + final velocity)(time interval)

Sample problem C page 53 A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and braking system, and comes to rest 5.5s later. Find the distance that the car travels during braking. vi=42 m/s vf=0.0 m/s Dt=5.5s Dx = 1/2 (vi + vf) Dt Dx = 1/2 (vi + vf) Dt = 0.5(42m/s+0m/s)(5.5s)=120m the calculator answer is 115.5 but round to two sig figs to get 120m. Do not use a decimal point so as to indicate the zero is NOT significant. Solve Practice C #1-5 page 53

Uniform Acceleration If final velocity is not known but initial velocity, uniform acceleration and elapsed time are known, the equations can be manipulated to solve for final velocity: vf = vi + aDt final velocity = initial velocity + (acceleration)(time interval) To determine displacement when only initial velocity, acceleration and time interval are known (final velocity is unknown): Dx = viDt + 1/2a(Dt)2

Sample problem D page 55 A plane starting at rest at one end of a runway undergoes a uniform acceleration of 4.8 m/s2 for 15s before takeoff. What is its speed at takeoff? How long must the runway be fore the plane to be able to take off: vi=0m/s a=4.8m/s2 Dt=15s vf= ? Dx= ? vf = vi + aDt vf= 0m/s + (4.8m/s2)(15s) = 72m/s Dx = viDt + 1/2a(Dt)2 Dx= (0m/s)(15s) + (0.5)(4.8m/s2)(15s)2 = 540m Solve Practice D #1-4 page 55

Uniform Acceleration If time is not known and final velocity is needed, if initial velocity, uniform acceleration and displacement are known, the equations can be manipulated to solve for final velocity: vf2 = vi2 + 2aDx final velocity squared = initial velocity squared + 2(acceleration)(displacement)

Sample problem E page 57 A person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500m/s2. What is the velocity of the stroller after it has traveled 4.75m? vi=0m/s a=0.500m/s2 Dx=4.75m vf= ? vf2 = vi2 + 2aDx vf2= (0m/s)2 + 2(0.500m/s2)(4.75m) = 4.75m2/s2 vf = +/-2.18m/s = +2.18m/s (moving in same direction it is accelerated) Solve Practice E #1-6 page 58