Question

The acceleration of a particle is given by a = 4t – 30, where a is in meters per second squared and t is in seconds. Determine the velocity and displacement as functions of time. The initial displacement at t = 0 is s0 = -5m, and the initial velocity is vo = 3m/s

Answer 1

Answer:

The velocity of the particle is given by $v(t) = 2\cdot t^{2}-30\cdot t + 3$, where $v$ is in meters per second and $t$ is in seconds.

The displacement of the particle is given by $s(t) = \frac{2}{3}\cdot t^{3}-15\cdot t^{2}+3\cdot t -5$, where $s$ is in meters and $t$ is in seconds.

Explanation:

The acceleration of the particle is given by $a(t) = 4\cdot t -30$, the expression for velocities are obtained by integration:

Velocity

$v(t) = \int {a(t)} \, dt$

$v(t) = \int {4\cdot t-30} \, dt$

$v(t) = 4\int {t} \, dt -30\int \, dt$

$v(t) = 2\cdot t^{2}-30\cdot t + v_{o}$

Where $v_{o}$ is the initial velocity of the particle, measured in meters per second.

If $t = 0$ and $v(0) = 3$, then:

$3 = 2\cdot (0)^{2}-30\cdot (0)+v_{o}$

$v_{o} = 3$

The velocity of the particle is given by $v(t) = 2\cdot t^{2}-30\cdot t + 3$, where $v$ is in meters per second and $t$ is in seconds.

Displacement

$s(t) = \int {v(t)} \, dt$

$s(t) = \int {2\cdot t^{2}-30\cdot t+3} \, dt$

$s(t) = 2\int {t^{2}} \, dt - 30\int {t} \, dt +3\int \, dt$

$s(t) = \frac{2}{3}\cdot t^{3}-15\cdot t^{2}+3\cdot t +s_{o}$

If $t = 0$ and $s(0) = -5$, then:

$-5 = \frac{2}{3}\cdot (0)^{3}-15\cdot (0)^{2}+3\cdot (0)+s_{o}$

$s_{o} = -5$

The displacement of the particle is given by $s(t) = \frac{2}{3}\cdot t^{3}-15\cdot t^{2}+3\cdot t -5$, where $s$ is in meters and $t$ is in seconds.