Question

The displacement of a particle in time t is given by s=2t^2-3t+1. The acceleration is?​

Answer 1

Answer:4m/s^2

Explanation:

Acceleration=dv/dt

V=ds/dt

V=d(2t^2-3t+1)/dt

=4t-3

A=d(4t-3)/dt

=4

Answer 2

We have,

Speed,

$v = \frac{d}{t} \\ = > v = \frac{2t ^{2} - 3t + 1 }{t}$

Again,

v=u + at

$\frac{2 {t}^{2} - 3t + 1}{t} = 0 + at \\ = > 2 {t}^{2} - 3t + 1 = a{t}^{2} \\ = > 2 {t}^{2} - 2t - t + 1 = a {t}^{2} \\ = > 2t( t - 1) - 1(t - 1) = a {t}^{2} \\ = > (2t - 1)(t - 1) = a {t}^{2} \\ = > a = \frac{(2t - 1)(t - 1)}{ {t}^{2} }$