Asked by vivian
A man drove his car north west for a distance of 10km,then east for 40km and then south for 60km.Calculate the overall displacement and direction of the car from the starting point.
Answers
Answered by
oobleck
the final location is <-10/√2,10/√2> + <40,0> + <0,-60>
So, evaluate the (x,y) value and then
displacement is √(x^2+y^2)
and the angle θ has tanθ = y/x
So, evaluate the (x,y) value and then
displacement is √(x^2+y^2)
and the angle θ has tanθ = y/x
Answered by
henry2,
All angles are measured CW from +y-axis.
AD = AB+BC+CD = 10km[315o] + 40km[90o] + 60km[180o],
AD=(10*sin315+40*sin90+60*sin180)+(10*cos315+40*cos90+60*cos180)I
AD = 32.93 - 52.93i = 62.3km[31.9o] S. of E. = 62.3km[121.9o] CW.
= Displacement.
AD = AB+BC+CD = 10km[315o] + 40km[90o] + 60km[180o],
AD=(10*sin315+40*sin90+60*sin180)+(10*cos315+40*cos90+60*cos180)I
AD = 32.93 - 52.93i = 62.3km[31.9o] S. of E. = 62.3km[121.9o] CW.
= Displacement.
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