a particle starts from the origin at t=0 with a velocity ^ of 10 j m/s and moves in the xy plane wid a constant ^ ^ acceleration of 8 i + 2j ms-2. (i) at what time is the x-coordinate of the particle 16m? what is y-coordinate of the particle at the time. (ii) what is the speed of the particle at that time.

Asked by  | 22nd Jul, 2010, 07:33: PM

Expert Answer:

u = uxi + uyj = 10j, ux = 0 and uy = 10 m/s
a = axi + ayj = 8i + 2j, ax = 8 and ay = 2 m/s2
x = uxt + (1/2)axt2
16 = 8 t2/2
t = 2 sec         ... Ans
y = uyt + ayt2/2  = 10x2 + 2x22/2 = 24 m
vx = ux + axt = 16 m/s
vy = uy + ayt = 14 m/s
Speed = √(vx2 + vy2) = 21.26 m/s
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TopperLearning.

Answered by  | 23rd Jul, 2010, 06:37: PM

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