Answer to Question #136622 in Mechanics | Relativity for Olaoluwa

Question

The motion of a particle is given by a =t^3-3t^2+5, where a is the acceleration in m/s^2 and t is the time in seconds .the velocity of the particle at t=1 seconds is 5.25m/s and the displacement and the velocity when the t =2seconds

Answer 1

Since the acceleration is $a = t ^ 3 - 3 \times t^ 2 + 5$ at $t = 1$ seconds and the velocity is at $t = 1$ seconds $v = 5.25$ m/s, the primary acceleration is the velocity, therefore the velocity is

$V (t) =\frac {t ^ 4} {4} -t ^ 3 +5 \times {t} + C$ , where $C$ is some factor.

We find from the data the coefficient $C$ , therefore $5.25 = \frac14-1 + 5 + C$ therefore $C=1$ , and therefore the initial speed at $t = 0$ is equal to $v = 1$ m/s,

The speed at $t = 2$ will be $v (2) =\frac {2 ^ 4} {4} -2 ^ 3 + 5\times {2} + 1 = 7$ m/s.

And by the path formula with equally accelerated motion, the path is $S = {v}\times {t} +\frac { {a}\times {t ^ 2}} {2}.$

Because $a = t ^ 3-3\times t ^ 2 + 5, S = {v}\times {t} +\frac { {t ^ 3- {3}\times {t} ^ 2 + 5} \times{t ^ 2}} {2}.$

$S = {7}\times {2} +\frac {{2 ^ 3- {3}\times {2} ^ 2 + 5} \times{2 ^ 2}} {2} = 16$ meters.