Question

The equation of motion of a particle is s = 2t3 − 8t2 + 4t + 2, where s is in meters and t is in seconds. (assume t ≥ 0.) (a) find the velocity and acceleration as functions of t.

Answer 1

The equation of motion of a particle is given by,

s = $2t^{3} - 8t^{2} + 4t +2$

Let velocity of particle is denoted by v.

The velocity of the particle is given by,

velocity (v) = $\frac{ds}{dt}$

v = $\frac{d}{dt} [ 2t^{3} - 8t^{2} + 4t +2 ]$

v = $\frac{d}{dt} 2t^{3} - \frac{d}{dt} 8t^{2} + \frac{d}{dt} 4t + \frac{d}{dt} 2$

v = $6t^{2} -16t +4$

Thus, Velocity of particle is given by,

v = $6t^{2} -16t +4$

Let acceleration of the particle is denoted by a.

The acceleration of particle is given by,

Acceleration = $\frac{dv}{dt}$

a = $\frac{d}{dt} [6t^{2} -16t +4]$

a = $\frac{d}{dt} 6t^{2} -\frac{d}{dt}16t +\frac{d}{dt}4$

a = 12t - 16

Thus, acceleration of the particle is given by,

a = 12t - 16