A particle moves in the xy-plane with constant acceleration. At time zero, the p

Question

A particle moves in the xy-plane with constant acceleration. At time zero, the particle is at x=6.50m, y=5.00 m, and has a velocity Vo = (3.00 m/s)i + (-7.50 m/s) j. The accerlation is given by a=(1.50 m/s^2) i + (-4.00 m/s^2)j. What is the x-component and y-component of velocity after 8 seconds? The solution offered is to use v=vo +at in each dimension separately. vx=(3.00m.s)+(8.5s)(1.50 m/s^2) and vy(7.50m/s)+(8.5s)(-4.00m/s^2). So v=(15.8 m/s)i +(-26.5 m/s)j. My question is, why does the Voy of -7.50 m/s turn positive when computing vy?

Explanation / Answer

? the general equation for velocity v(t)=v0 +a*t, where v0=(2, -3), a=(4, 3), t=8.5; thus v(t) = (2, -3) +(4, 3)*8.5 = (2+4*8.5, -3+3*8.5) = = (36, 22.5); vx= 36m/s, vy=22.5m/s; ? the general equation for position r=r0+v0*t +0.5*a*t^2; r=(5, 6) +(2, -3)*8.5 +0.5*(4, 3)*8.5^2 = = (5, 6) +(17, -25.5) +(144.5, 108.375) = = (5 +17 +144.5, 6 –25.5 +108.375) = (166.5, 88.875); rx=0.1665 km, ry=0.088875 km;

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