 # Kinematics- Acceleration Chapter 5 (pg. 81-103) A Mathematical Model of Motion.

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Kinematics- Acceleration Chapter 5 (pg. 81-103) A Mathematical Model of Motion

The Nature of Acceleration  Acceleration is the rate of change in _______ with time - the rate at which the velocity changes. If an object is speeding up, it is __________ ! If an object is slowing down, it is ! If an object has constant speed, but is changing direction, it is ! -this is the case when an object travels in a circular path at constant speed ANY VELOCITY CHANGE is an ACCELERATION ! velocity accelerating

Calculating Acceleration Acceleration is a vector, so in a straight line the direction will be represented by +/- sign.  Negative acceleration means acceleration in the opposite direction of whatever positive means!  Average acceleration is change in velocity divided by change in time.  Uniform acceleration means constant acceleration!

a av = v - v o ∆t This is the basic definition of acceleration in the form of an equation. The following equations are basic equations which are derivations of our basic definitions of velocity and acceleration for objects moving with constant acceleration: m/s 2 are the units of acceleration ( think meters per second CHANGE in velocity per second of time) v = v 0 + at ∆d = v 0 t + 1212 at2at2 ∆d = (v 0 + v) 1212

A car accelerates from rest at a rate of 2.00 m/s 2 west for 8.50 s. What is the velocity of the car at this time? What was its displacement during this time? v 0 = 0m/s a = 2.00 m/s 2 west ∆t = 8.50 s v = ? ∆d = ? v = v 0 + at v = 0 + (2.00 m/s 2 west )(8.50 s) v = 17.0 m/s west ∆d = 0 + 0.5(2.00 m/s 2 west)(8.50s) 2 ∆d = 72.3 m west ∆ d = v 0 t + a t 2 1212

3) An object is accelerated to 35.0 m/s at a rate of 2.00 m/s 2 for a time of 7.34 s. How far did it move during this time? 1) How far does an object travel if it accelerates from 12.0 m/s to 45.0 m/s in 15.3 s? 2) An object traveling 48.7 m/s is brought to a stop in 7.61 s. How far did the object move during this time? 4) An object rolled up a hill will stop after having traveled up the hill 18.0 m. If the object is slowed at a rate of 2.50 m/s 2, how long will it take to stop rolling? [203m] [436m] [185m] [3.80s]

Basic Equations: v = v 0 + at ∆d = (v 0 + v) 1212 ∆d = v 0 t + 1212 at2at2

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