a particle moves in the x-y plane with constant acceleration. At time t=0s, the

Question

a particle moves in the x-y plane with constant acceleration. At time t=0s, the position vector for the particle is d=4.8mx+10my. The acceleration is given by the vector a=7.5m/s2x+5.8m/s2y. The velocity at time t=0 is v=7.8m/sx-2.6m/sy.
a) Find the magnitude of the velocity vector at t=3.4s.
b). What's the angle between the velocity vector and the x-axis at t=3.4s
c). What's the magnitude of the position vector at t=3.4s
d). what's the angle between the velocity vector and the x-axis at t=3.4s

can you give me formulas i'm supposed to use and why please?

Explanation / Answer

at t=3.4
v=(7.8+7.5*3.4)x+(-2.6+5.8*3.4)y =33.3x+17.2y
d= (4.8+ 7.8*3.4+.5*7.5*3.4^2)x+(10-2.6*3.4+.5*5.8*3.4^2)y = 74.67x +34.684y
a) 37.48m/s
b) 27.320
c) 82.33m
d) 24.910

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