1.At time t = 0 s, an object is observed at x = 0 m; and its position along the

Question

1.At time t = 0 s, an object is observed at x = 0 m; and its position along the x axis follows this expression: x = –3t +t3, where the units for distance and time are meters and seconds, respectively.What is the object’s displacement Dx between t = 1.0 s and t = 3.0 s?

2.An elevator is moving upward with a speed of 11 m/s. Three seconds later, the elevator is still moving upward, but its speed has been reduced to 5.0 m/s. What is the average acceleration of the elevator during the 3.0 s interval? upward or dawnward?

Explanation / Answer

Given 1.x = -3t+t^3 At t = 1.0s, x = -3(1 s)+(1 s)^3    = -2 m At t = 3.0 s, x= -3(3.0 s)+(3.0 s)^3    = 18 m
Displacement, Dx = 18 m - (- 2 m)
= 20 m ------------------------------------------------------------------------- 2.Initial velocity , vi = 11 m/s
After 3 s, the velocity is vf = 5.0 m/s Average acceleration = (vf-vi)/t    = 5-11/3    = -2 m/s^2
The elevator is moving in downward direction

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