Question

A particle starts from the origin at t = 0 s with a velocity of 10.0 j m/s and moves in the x-y plane with a constant acceleration of (8.0i + 2.0j) m s⁻².a. At what time is the x-coordinate of the particle 16m? What is the y-coordinate of the particle at that time?b. What is the speed of the particle at the time

in this question why is ux 0​

Answer 1

Answer:

21.26m/s

here ux is 0 because the initial velocity is 0 as mentioned in question

Explanation:

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