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

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

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

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

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

1 1 ∆x = (v + v0)(v – v0) = (v2 – v0 2) 2a 2a 2a∆x = v2 – v0 2 4 v2 = v0 2 + 2a∆x Present: v, a, ∆x Specifically: v2 = v0 2 + 2a(x – x0)

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 = v0 2 + 2a(x – x0) x, v, a