Question
A particle's constant acceleration is north at 108 m/s2. At t = 0, its velocity vector is 61.7 m/s east. At what time will the magnitude of the velocity be 102 m/s?
Answers
drwls
The east velocity component remains Vx = 61.7 m/s, since the acceleration is in a perpendicular direction.
When the magnitude of the velocity is 102 m/s, the north component must be
Vy = sqrt[102^2 - 61.7^2] = 81.2 m/s
The time required to attain this value of Vy is
t = (81.2 m/s)/(108 m/s^2) = 0.752 s
When the magnitude of the velocity is 102 m/s, the north component must be
Vy = sqrt[102^2 - 61.7^2] = 81.2 m/s
The time required to attain this value of Vy is
t = (81.2 m/s)/(108 m/s^2) = 0.752 s