Answer:
5
Explanation:
To solve with calculus, distance is the integral of speed:
d = ∫ |v| dt
d = ∫₀⁴ |t − 3| dt
d = -∫₀³ (t − 3) dt + ∫₃⁴ (t − 3) dt
d = ∫₀³ (3 − t) dt + ∫₃⁴ (t − 3) dt
d = (3t − ½ t²) |₀³ + (½ t² − 3t) |₃⁴
d = [ (9 − 9/2 ) − (0 − 0) ] + [ (8 − 12) − (9/2 − 9) ]
d = 9/2 + 1/2
d = 5
You can also find this geometrically. Graph y = |x − 3|, then find the area under the curve. You will find it's the area of two triangles.
d = ½ (3)(3) + ½ (1)(1)
d = 5
It's important to note that distance is not the same thing as displacement. Displacement is the difference between where you start and where you stop. Distance is length of the path you take.