# What is the magnitude of the displacement from the origin (x=0 m, y=0 m) after 4s? A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x=6.5m, y=2.5m, and has velocity Vo= (4m/s)i + (-4m/s)j. The acceleration is given by a=(1.5m/s^2)i+ (1.5m/s^2)j. What is the magnitude of the displacement from the origin (x=0 m, y=0 m) after 4s? Answer in units of m The initial position of the particle is (6.5, 2.5). It has an initial velocity of Vo= (4m/s)i + (-4m/s)j and the acceleration is given by a=(1.5m/s^2)i+ (1.5m/s^2)j

Use the formula s = u*t + (1/2)*a*t^2 where s is the distance traveled, u is the initial velocity, t is the time and a is the constant acceleration.

This gives the x-component of the displacement after 4 s as X = 4*4 + (1/2)(1.5)(4)^2 = 16 + 12 = 28 m. The y-component of the displacement is Y = (-4)*4 + (1/2)*(1.5)*4^2 = -4

The position after 4s is (6.5 + 28, 2.5 - 4) = (34.5 , -1.5)

The magnitude of the displacement from the origin is sqrt(34.5^2+1.5^2) - sqrt(6.5^2+2.5^2)

=> sqrt(1192.5) - sqrt(48.5)

=> 27.56 m

The magnitude of the displacement from the origin in 4 s is approximately 27.56 m

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