Question

x=2t³-21t²+60t+6 find the value of acceleration of the particle when velocity is 0

Answer 1

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