Question

position x of a particle moving aling straight line is related with time t as x=(t^3-3t^2+2t-7) m where t is in second the velocity of particle when Acceleration becomes 0​

Answer 1

Answer:

velocity = -1 m/ s

Step-by-step explanation:

$x = {t}^{3} - 3 {t}^{2} + 2t - 7 \\ \frac{dx}{dt} = 3 {t}^{2} - 6t + 2 \\ therefore \: \: \: v = 3 {t}^{2} - 6t + 2 \\ \\ \frac{dv}{dt} = 6t - 6 \\ \\ a = 6t - 6 \\ \\ when \: \: \: a = 0 \\ \\ then \: \: \: \: \: 6t - 6 = 0 \\ t = 1 \: s \\ \\ then \: \: \: \: \: \: \: v = 3 {(1)}^{2} - 6(1) + 2 \\ \\ required \: velocity \: \: v = 3 - 6 + 2 = - 1 \: \: \: m {s}^{ - 1}$