Question

John is rollerblading down a long straight path. At time zero, there is a mailbox about 1 m in front of him. In the 5-second time period that follows, John's velocity is given by the right-hand graph in the figure below. Taking the mailbox to mark the zero location, with positions beyond the mailbox as positive, plot his position versus time in the left-hand graph below.  

Assuming that all the numbers given above are exact, what is John's position at a time of 4.23 seconds? Enter your answer to at least three significant digits.

Answer 1

New position = previous position + ( velocity x duration)

X’ = x + (v Δt)

X’ = x + v (T₂ - T₁)

Time (s)	John’s Position (m)	Velocity (between T1 and T2) (m/s)
0	= -1 ( because john is behind the mailbox)	NA
1	= -1 + (-2 x1) = -3	-2 ( between 0s and 1s)
2	= -3 + (0 x 1) = -3	0 ( between 1s and 2s)
3	= -3 + (2 x1) = -1	2 ( between 2s and 3s)
4	= -1 + (2x1) = 1	2 ( between 3s and 4s)
5	= 1 + (2x1) = 3	2 ( between 4s and 5s)

At 4s

x = 1 m

At 4.23 s

X’ = x + v (4.23-4)

X’ = 1+ 2(0.23)

X’ = 1.46 m

so at 4.23 s John's position is 1.46 m