A particle leaves the origin with initial velocityv 0 = 8i+ 22j m/s. It undergoe

Question

A particle leaves the origin with initial velocityv0 = 8i+ 22j m/s. It undergoes a constant accelerationgiven by a = –1.7i +0.27j m/s2 (a) When [i.e., for t greater than zero] does theparticle cross the y axis?
t = 1 s (b) What is its y coordinate at that time?
y = 2 m

(c) How fast is it moving, and in what direction, at that time?
v = 3 m/s
v = 4° from the +xaxis

A particle leaves the origin with initial velocityv0 = 8i+ 22j m/s. It undergoes a constant accelerationgiven by a = –1.7i +0.27j m/s2 (a) When [i.e., for t greater than zero] does theparticle cross the y axis?
t = 1 s (b) What is its y coordinate at that time?
y = 2 m

(c) How fast is it moving, and in what direction, at that time?
v = 3 m/s
v = 4° from the +xaxis

Explanation / Answer

v0 = 8i + 22j m/s. a = –1.7i + 0.27jm/s2 v(t) = a dt + C =-1.7 t i + 0.27 t j + 8 i + 22 j x(t) = vx dt + C1 = -0.85 t2 + 8 t y(t) = vy dt + C2 = 0.135 t2 + 22 t a) x = 0, -0.85 t2 + 8 t = 0, t = 8/0.85 = 9.41 s b) y(9.41) = 219 m c) v(9.41) = -8.0 i + 24.5 j v = 25.8 m/s v = 360 - tan-1(24.5/8.0) =288o

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