Asked by jacob
Jim is 100m(S) of Paul. Paul begins to walk east at 2.0m/s the same time Jim begins to walk west at 3.0m/s. what is the displacement of Jim from Paul 20 seconds later?
Please explain steps
Answer is 141(SW)
Please explain steps
Answer is 141(SW)
Answers
Answered by
Henry
A = 3m/s * 20s = 60m[180o].
B = 100m[270o].
C = 2m/s * 20s = 40 m[0o].
D = (A+B)-C
A+B = (60*cos180+100*cos270)i(60*sin180+
100*sin270)
A+B = (-60+0) + i(0-100)
A+B = -60 - i100 = -116.5[59o]
A+B = 116.5[59+180] = 116.5m[239o].
D = (A+B)-C = 116.5[239] - 40
D = -60 - i100 - 40 = -100 - i100
D = -141.4[45o[ = 141.4[45+180]
D = 141.4m[225o] = 141.4m[SW].
B = 100m[270o].
C = 2m/s * 20s = 40 m[0o].
D = (A+B)-C
A+B = (60*cos180+100*cos270)i(60*sin180+
100*sin270)
A+B = (-60+0) + i(0-100)
A+B = -60 - i100 = -116.5[59o]
A+B = 116.5[59+180] = 116.5m[239o].
D = (A+B)-C = 116.5[239] - 40
D = -60 - i100 - 40 = -100 - i100
D = -141.4[45o[ = 141.4[45+180]
D = 141.4m[225o] = 141.4m[SW].
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