Asked by Yihalem Tewachew
To get to school a girl walks 1 km North in 15 minutes she th
en walks 200metere Southwest in 160 second find her average velocity
en walks 200metere Southwest in 160 second find her average velocity
Answers
Answered by
Damon
well, velocity is a vector so we need to find her displacement vector from start to finish.
1 km in y direction
then .2 cos 45 in negative y direction and .2 sin 45 in negative x direction
so in the end
y = 1 - .141 = .858 km north
x = -.141 or .141 km west
distance from start =sqrt(.858^2+.141^2) = sqrt(.736+.020) =.869 km
time = .25 hr + 160/3600 hr = .25 + .04 = .29 hr
so .869 km/.29 hr = 3 km/hour
direction = tan^-1 (.141/.858) degrees west of north
or 9.23 deg west of north which is 360-9.23 =351 on your compass if all directions given are magnetic not true.
1 km in y direction
then .2 cos 45 in negative y direction and .2 sin 45 in negative x direction
so in the end
y = 1 - .141 = .858 km north
x = -.141 or .141 km west
distance from start =sqrt(.858^2+.141^2) = sqrt(.736+.020) =.869 km
time = .25 hr + 160/3600 hr = .25 + .04 = .29 hr
so .869 km/.29 hr = 3 km/hour
direction = tan^-1 (.141/.858) degrees west of north
or 9.23 deg west of north which is 360-9.23 =351 on your compass if all directions given are magnetic not true.
Answered by
henry2,
All angles are measured CW from +y-axis.
V1 = 1000m/900s = 1.11 m/s[0o].
V2 = 200m/160s = 1.25 m/s[225o].
V = 1.11[0o] + 1.25[225o].
V = (1.11*sin0+1.25*sin225) + (1.11*cos0+1.25*cos225)I,
V = -0.884 + 0.226i = 0.912m/s[-90o] = 0.912m/s[270o] CW.
V1 = 1000m/900s = 1.11 m/s[0o].
V2 = 200m/160s = 1.25 m/s[225o].
V = 1.11[0o] + 1.25[225o].
V = (1.11*sin0+1.25*sin225) + (1.11*cos0+1.25*cos225)I,
V = -0.884 + 0.226i = 0.912m/s[-90o] = 0.912m/s[270o] CW.
Answered by
Habtamu
Answer
Answered by
Rebuma
graphh is important for these questio
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