# 3 - 2.4 Kinematics for Constant Acceleration Now we can...

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2.4 Kinematics for Constant Acceleration Now we can apply the concepts of displacement , velocity , and acceleration to describe the motion of objects. Before we begin, let’s make one more simplifying condition – We will only deal with objects that undergo constant acceleration . Of course, it’s quite possible that an object’s acceleration is also changing in time. Question : Suppose an object starts from rest and undergoes constant acceleration, a , for some time t . How far does the object go in that time? Well, t v v a o f ± Since the object starts from rest, its initial velocity is zero! Thus, the final velocity is just, at v f Since I now know the object’s initial and final velocities, I can calculate its average velocity: 2 f o v v v ² at v f 2 1 2 1

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I know the relationship between average velocity and distance traveled: * ' t x v * at 2 1 Bring the t out of the denominator, and we find: 2 1 a * * ' 2 t x *This is the distance an object travels when it starts from rest and moves with constant acceleration a for time constant acceleration, , for time t . Example : How far have I gone in my car if I accelerate from rest at a constant 10.0 m/s 2 for 3.00 s? 2 2 1 at x ' 2 2 2 1 s) 00 . 3 )( m/s 0 . 10 ( m 0 . 45 Notice how the units cancel just right to give me m (a distance). So, gives the distance for an object starting from rest. 2 2 1 t a x * * ' But what if it’s already moving when it starts to accelerate, i.e. its initial velocity is no longer zero???
Example : Let’s say I’m driving along in my car at a constant 27.8 m/s, and then I begin to accelerate at 10.0 m/s 2 for 3.00 s. Now how far does my car go in those 3.00 s? Well, we just calculated what the distance traveled is going to be due to the acceleration: 2 2 1 at x ' 2 2 2 1 s) 00 . 3 )( m/s 0 . 10 ( m 0 . 45 So the car will travel 45 m in those 3 s just due to the acceleration alone. But what about the initial velocity? Surely the car covers some distance since it was already moving since it was already moving. Well, from the definition of velocity, we know that: t x v ' So, due to the velocity is just: vt x ' x ' m 83.4 s) m/s)(3.00 8 . 27 ( Thus, the total distance covered = Distance due to acceleration + Distance due to initial velocity = 45.0 m + 83.4 m = 128 m 2 * * * In general then,

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## This note was uploaded on 03/17/2011 for the course PHYSICS 2001 taught by Professor Young during the Spring '08 term at LSU.

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3 - 2.4 Kinematics for Constant Acceleration Now we can...

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