x=2t³-21t²+60t+6 find the value of acceleration of the particle when velocity is 0
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Answer:
a = -18 m/s^2 , 18 m/s^2
Explanation:
x = 2t^3 - 21t^2 + 60t + 6
v = dx/dt = d(2t^3 - 21t^2 + 60t + 6)/dt
v = 6t^2 - 42t + 60
According to question,
v = 0
6t^2 - 42t + 60 = 0
6(t^2 - 7t + 10) = 0
t^2 - 7t + 10 = 0
t^2 - 5t - 2t + 10 = 0
t(t - 5) -2(t - 5) = 0
(t - 2)(t - 5) = 0
At t= 2 & 5s v = 0
a = dv/dt = d(6t^2 - 42t + 60)/dt
a = 12t - 42
At t = 2s
a = 12 × 2 - 42
a = 24 - 42
a = -18m/s^2
At t = 5s
a = 12 × 5 - 42
a = 60 - 42
a = 18m/s^2
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