a=(5.97 @ (-50o)-5.30 @ 35o)/t.
a = ((3.84-4.57i) - (4.34+3.04i))/2
a = (3.84-4.57i -4.34-3.04i)/2
a = (-0.5 - 7.61i)/2
a = -7.63 @ 86.2o/2 = -3.82m/s^2 @ 86.2o
X = -3.82*cos86.2 = -0.253 m/s^2.
Y = -3.82*sin86.2 = -3.81 m/s^2.
Initially, a particle is moving at 5.30 m/s at an angle of 35.0° above the horizontal (+x axis). Two seconds later, its velocity is 5.97 m/s at an angle of 50.0° below the horizontal. What was the particle's average acceleration during this time interval?
________m/s2 (x component)
________m/s2 (y component)
1 answer
1 answer