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Question
For the one-dimensional motion, described by x = t – sint
- x (t) > 0 for all t > 0.
- v (t) > 0 for all t > 0.
- a (t) > 0 for all t > 0.
- v (t) lies between 0 and 2.
Solution
a and d
Explanation:
Given, `x = t - sin t`
Velocity v = `(dx)/(dt) = d/(dt) [t - sin t]`
= `1 - cos t`
Acceleration a = `(dv)/(dt) = d/(dt) [1 - cos t] = sin t`
As acceleration a > 0 for all t > 0
Hence, x(t) > 0 for all t > 0
Velocity v = `1 - cos t`
When, `cos t = 1`, velocity v = 0
vmax = `1 - (cos t)_"min" = 1 - (- 1) = 2`
vmin = `1 - (cos t)_"max" = 1 - 1 = 0`
Hence, v lies between 0 and 2.
Acceleration a = `(dv)/(dt) = - sin t`
When t = 0; x = 0, x = + 1, a = 0
When t = `π/2`; x = 1, v = 0, a = – 1
When t = π; x = 0, x = – 1, a = 1
When t = 2π; x = 0, x = 0, a = 0
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