Question

y=2t^3-3t^2+5 find the position velocity and acceleration of the particle at the end of 2s. here y is measured in metre and t in seconds.Plz help me I will make you brainliest
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Answer 1

Given :-

Position of Particle = y = 2t³ - 3t² + 5

Time = 2s

We are asked to find Velocity and Acceleration at the end of 2 s. Here, we will differentiate.

V = dy/dt

V = d(2t³ - 3t² + 5)/dt

V = 6t² - 6t

Putting t = 2s

V = 6(2)² - 6(2)

V = 24 - 12

V = 12 ms-¹

For Acceleration, Differentiating 'V' w.r.t 't'.

a = dv/dt

a = d(6t² - 6t)/dt

a = 12t - 6

Putting t = 2s

a = 12(2) - 6

a = 18 ms-²

Hence,

Velocity of Particle = V = 12 ms-¹

Acceleration of Particle = a = 18 ms-².