An object is at x 0 at t 0 and moves along the x ax s according to the velocity-

Question

An object is at x 0 at t 0 and moves along the x ax s according to the velocity-time graph shown below. ve (m/s) 20 EEEEEEEEEEEEEEEEENE -10 (a) What is the object's acceleration between 0 and 4.0 s? m/s2 (b) What is the object's acceleration between 4.0 s and 9.0 s? m/s (c) WI is the object's acceleration between 13.0 s and 18.0 s hat m/s (d) At what time(s) is the object moving with the lowest speed? (If there is more than one time, enter the times from smallest to largest. If there is only one time, enter 0 in the second box (e) t what time is the object farthest from x 0? (f) What is the final position x of the object at t 18 s g) Through what total distance has the object moved between 0 and t 18 s

Explanation / Answer

a)a=0(slope=0)

b)a=(19-(-11))/5=6m/s2

c)a=(0-19)/5=-3.8m/s

d)6s

18s

e)The object is farthest when net area under curve is greatest.

t=18s

f)x=area under curve=-(10*4+0.5*10*2)+(0.5*19*3+19*4+0.5*19*5)=102m

g)d=10*4+0.5*10*2+0.5*19*3+19*4+0.5*19*5=202m

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