A body starts from origin moves along the x axis such that the velocity at any instant is given by( 4t^3-2t) .what is the acceleration of the particle when it is at a distance 2 meter from the origin?
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Answered by
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v = 4t³ - 2t
a = dv/dt = 12t² - 2 ---------------------------(1)
ds/dt = v = 4t³ - 2t
ds = (4t³ - 2t)dt
integrate both side limit of s is 0 to 2 & limit of t is 0 to t
2 = t^4 - t²
t^4 - 2t² + t² - 2 = 0
t²(t² - 2) + 1(t² - 2) = 0
(t² + 1)(t² - 2) = 0
t² - 2 = 0
t = √2 (as time can not be negative so left t = -√2)
t² + 1 = 0 (never zero)
put t = √2 in equation (1)
a = 24 - 2
a = 22 m/s²
a = dv/dt = 12t² - 2 ---------------------------(1)
ds/dt = v = 4t³ - 2t
ds = (4t³ - 2t)dt
integrate both side limit of s is 0 to 2 & limit of t is 0 to t
2 = t^4 - t²
t^4 - 2t² + t² - 2 = 0
t²(t² - 2) + 1(t² - 2) = 0
(t² + 1)(t² - 2) = 0
t² - 2 = 0
t = √2 (as time can not be negative so left t = -√2)
t² + 1 = 0 (never zero)
put t = √2 in equation (1)
a = 24 - 2
a = 22 m/s²
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