Kinematics Unit 2 B Motion Part 5 Kinematics

Kinematics Unit 2 B: Motion – Part 5

Kinematics • Kinematics equations take all of the simple motion concepts we’ve discussed and combine them into a few more complex equations.

Equations 1. xf = ½at 2 + vit +xi 2. Δx = ½at 2 + vit 3. vf = at + vi 4. vf 2 = vi 2 + 2 aΔx 5. Δx = ½t(vi+vf) • xf – final position • xi – initial position • Δx – displacement/change in position • a – acceleration • t – time • vi – initial velocity • vf – final velocity

Example 1 A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and braking system, and come to rest 5. 5 s later. Find the distance the car travels during breaking. vi=42 m/s vf=0 m/s t=5. 5 s Δx=? Δx = ½t(vi+vf) Δx = ½(5. 5)(42+0) Δx = ½(5. 5)(42) Δx = 115. 5 m

Example 2 A plane starting at rest at one end of a runway undergoes a uniform acceleration of 4. 8 m/s 2 for 15 s before takeoff. What is the speed at takeoff? How long must the runway be for the plane to be able to take off? a=4. 8 m/s 2 t=15 s vi=0 m/s vf=? Δx=? vf = at + vi vf 2 = vi 2 + 2 aΔx vf = (4. 8)(15)+0 vf 2 -vi 2=2 aΔx vf = 72 m/s vf 2 -vi 2/2 a=Δx 722 -02/2(4. 8)=Δx 5184/9. 6=Δx 540 m=Δx

Example 3 A person pushing a stroller starts from rest, uniformly accelerating at a rate of 0. 500 m/s 2. What is the velocity of the stroller after it has traveled 4. 75 m? vi=0 m/s a=0. 500 m/s 2 vf=? Δx=4. 75 m vf 2 = vi 2 + 2 aΔx vf 2 = 02 + 2(. 5)(4. 75) √vf 2 = √ 4. 75 vf = 2. 18 m/s