1. Motion with constant acceleration
AP Physics
Section 2-5
Motion with constant acceleration

2. Velocity, Acceleration & Time
— If you have no displacement information.
To find final velocity, we rearrange equation B: – ∆v v v a B = = ∆t ∆t
v – v0
Assume: a =
a is always average, so
a = a. t
v – v0 = at
tinitial = 0, so ∆t = t v v0 2 = + a t
This equation is .
Present: v, a, t
linear

3. Position with Average Acceleration
Displacement, Acceleration & Time.
From equation 1 & 2:
Substitute 2 into 1. 1
∆x = ½ (v + v0) t 2 v = v0 + a t
∆x = ½ [(v0 + at)+ v0 ] t
∆x = ½ (2v0 + at) t
quadratic equation 3
∆x = v0 t + ½at2
Simple form, final position:
x = x0 + v0 t + ½at2
Present: ∆x, a, t

4. Under gravity, up positive:
Uses for equation 3
Equation 3 is often used where final velocity is hard to determine.
Equation 3 is often used where final velocity is hard to determine.
One common use is for falling objects, where the velocity of a projectile hitting the ground is not known.
One common use is for falling objects, where the velocity of a projectile hitting the ground is not known.
One common use is for falling objects, where the velocity of a projectile hitting the ground is not known.
In this case of vertical motion, the x is often replaced by y. We must specify v0 in the y direction.
In this case of vertical motion, the x is often replaced by y. We must specify v0 in the y direction.
In this case of vertical motion, the x is often replaced by y. We must specify v0 in the y direction. 3
∆y = v0 y t + ½at2
Under gravity, up positive:
For falling objects, a = –g
y = y0 + v0 y t – ½gt2

5. Displacement, Velocity & Acceleration
— If you have no time information.
From equation 1 & 2:
Solve equation 2 for t. a t v v – v0 2 = v0 + a t = v – v0
Substitute t into 1. t = a 1
∆x = ½ (v + v0) t v – v0
∆x = ½ (v + v0) a

6. 1 1 ∆x = (v + v0)(v – v0) = (v2 – v0 2) 2a 2a 2a∆x = v2 – v0 24
v2 = v a∆x
Present: v, a, ∆x
Specifically:
v2 = v a(x – x0)

7. Four Kinematic Equations
The four kinematic equations if t0 = 0:
variables
x = x0 + ½ (v + v0)t
x, v, t 1
v, a, t 2
v = v0 + at
x = x0 + v0 t + ½at2
x, a, t 3 4
v2 = v a(x – x0)
x, v, a