Question
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
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
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.