Question

A car starts from xi = 12 m at ti = 0 and moves with the velocity graph shown in figure below.  

Part A

What is the object's position at t = 2 s? 

Part B

 What is the object's position at t = 3 s? 

Part C

What is the object's position at t = 4 s?  

Answer 1

part A )

X(t= 2sec) = 12m +(2*4 +8*.5*2)

X(t=2sec) = 12+16

X(t=2sec) = 28 m

Part B)

X(t=3sec) = 12+(0.5*3*12)

X(t=3sec) = 30m

Part C )

X (t=4) = 12+(0.5*3*12 -0.5*1*4)

X(t=4sec) = 28m

Answer 2

I know that t=2 is 36 m. This is because if you find the area of the triangle under t=2, you get an area of 8. There is alos a remaining rectangle that also has a area of 8. together you get 16, plus the starting 20 m.

Solve t=3 the same way. Not sure about t=4 yet, but the car changes directions at t=3, at 0m/s.