A particle moves along the x axis. It is initially at the position 0.130 m, movi

Question

A particle moves along the x axis. It is initially at the position 0.130 m, moving with velocity 0.050 m/s and acceleration -0.450 m/s2. Suppose it moves with constant acceleration for 5.50 s. (a) Find the position of the particle after this time. m (b) Find its velocity at the end of this time interval. m/s Next, assume it moves with simple harmonic motion for 5.50 s and x = 0 is its equilibrium position. (Assume that the velocity and acceleration is the same as in parts (a) and (b).) (c) Find its position m (d) Find its velocity at the end of this time interval. m/s

Explanation / Answer

a)use s=u*t+1/2a*t^2, we get s=-6.5, final position= -6.53+.13=-6.4m b)use v=u+a*t, we get v=-2.425m/sec meaning the direction is reversed

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