Question

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 Δx between t = 1.0 s and t = 3.0 s? *​

Answer 1

Given : 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

To Find : object’s displacement Δx between t = 1.0 s and t = 3.0 s?

Solution:

x = -3t + t³

at t = 0

x = -3(0) + (0)³ = 0

at 0 sec , 0 m

at t = 1

x = - 3(1) + 1³

=> x = - 2

at 1 sec   x = - 2 m

at t = 3

x = -3(3) + 3³

=> x =   18

t =  3 sec x =   18 m

displacement Δx between t = 1.0 s and t = 3.0 s

= 18 - ( - 2)

= 20  m

object’s displacement Δx between t = 1.0 s and t = 3.0 s = 20 m

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