Deriving Equations

(3) d = v1Δt + ½aΔt²

Derive from (2) v2= v1 + at

Substitute V2 to V1 to (1) d = 1/2 (v1+v2) t

(4) d = v2Δt - ½aΔt²

Deive from (2) v1= v2 - at

Substitue V1 to V2 to (1)

(5)v2² = v1² + 2ad

Multiply (1) and (2)

(3) d = v1Δt + ½aΔt²
d= area of the rectangle + area of the triangle     
= v1Δt + ½(at)Δt                 
= v1Δt + ½aΔt²
(4) d = v2Δt - ½aΔt²

d = area of the large rectangle - area of the triangle

= v2Δt  - ½(at)Δt

= v2Δt - ½aΔt²

• This is d,t graph
• From the starting point , I moved away from the starting point [E] by constant speed
• Then, i stopped for a while and walked towards the starting point [W] for few secods
• Then, stopped for the rest of motion
• 
  

• This is v,t graph
• The velocity was 0 for 2 seconds , which means the stopped motion
• The velocity was 0.5 after 2 seconds until 5 seconds ; walking motion [E]
• The velocity was 0 at 5 seconds until 7 seconds; no motion
• The velocity was -0.5 until ends ; walking motion [w]

• v,t graph
• velocity was increased 0 to 0.5 until 4 sec; speeding motion [E]
• From 4 sec the velocity was constant until 6sec ; constant speed [E]
• From 6 the velocity was -0.4 until 9 sec ; slower speed [w]
• Then, the motion stopped from 9sec

• d,t graph
• Started from 1 m , moved constant speed until 3sec ; moving [E]
• 3 to 7 sec rest motion 
• 7 to 10 ; moving constant speed[E] 

• v,t graph
• speed [E]was constant until 3 sec and 4 sec to 7 sec the speed was [W] constant then the motion stopped from 7 sec

• This is d , t graph 
• started from 3 m , constant speed [W]
• 3 second rest motion until 4 sec
• 4 sec constant speed [W] until 5 sec
• 5 sec rest motion until 7 sec
• 7 sec to 10 sec constant speed [E] to 3m