1. V = a*t = 1.8m/s^2 * 4.5s. = 8.1 m/s.
X = 6.1 m/s.
Y = 8.1 m/s.
tan A = Y/X = 8.1/6.1 = 1.32787
A = 53o = Direction.
2. V = Y/sinA = 8.1/sin53 = 10.14 m/s.
A particle travels to the right at a constant
rate of 6.1 m/s. It suddenly is given a vertical
acceleration of 1.8 m/s
2
for 4.5 s.
What is its direction of travel after the
acceleration with respect to the horizontal?
Answer between −180◦
and +180◦
.
Answer in units of ◦
What is the speed at this time?
Answer in units of m/s
1 answer