2. (20 pts) A car, facing in the +x direction, moves along the x-axis according

Question

2. (20 pts) A car, facing in the +x direction, moves along the x-axis according to the vélocity vs. time shown below 15 Slop -acceleaha s) is the car B. 10 s to 15 s wards (ie, in reverse)? Circle all that 5 s to 20 s a. (5 pts) During which A. 0 to J8 to 30s 30 to 35 F.35s to 40 For your following calculations, assume at least 2 significant figures on all values b. (10 pts) Carefully and accurately, graph the car's acceleration vs. time on the axes provided. (Show your work on the back of this (7.5) is (o) ime lo 0%" -337 c. (5 pts) Calculate the car's displacement at t- 40s.Show your work Page 2/8

Explanation / Answer

For moving backwards, the curve needs to be under the x-axis

So, the intervals are as follows :

i) 15-20 s

ii) 20-30 s

iii) 30-35 s

b)

acceleration is given by the slope of the velocity vs time curve.

So, the graph will look as follows :

during t = 0 to t = 10s, it will be a straight horizontal line with height = 15/10 = -1.5 m/s2 <---- negative

during t = 10s to t = 15s, the height will be 0 (along the x-axis)

during t = 15s to t=20s, the height will be : -5/5 = -1 m/s2

during t = 20s to t = 30s, the height will be again 0

during t = 30s to t = 35s, the height will be = 1 m/s2

during t = 35s to t = 40s, the height = 0

c)

displacement = total area under the curve of velocity vs time

area under the curve between t = 0 tot t = 10s : A1 = 0.5*10*15 = 75 m

area under the curve between t = 15 s to t = 35 s, A2 = -(2*0.5*5*5 + 5*10) = -75 m

so, total displacement = A1 + A2 = 75 - 75 = 0

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