Question

The acceleration of a particle is given by a = 4
t - 30, where a is in m/s2 and t is in seconds.
Determine the velocity and displacement as
functions of time. The initial position at t = 0 is
-5m and the initial velocity is vo = 3m/s.​

Answer 1

Answer: refer the material.

Explanation:

Answer 2

Answer:

Velocity, $v(t) = (2t^2-30t+3)\ m/s.$

Displacement, $x(t)=\left (\dfrac{2t^3}3-15t^2+3t-5\right )\ m.$.

Explanation:

Given:

• Acceleration of the particle, $a=4t-30$.

• Initial position of the particle, $x_o = -5\ m.$

• Final position of the particle, $v_o = 3\ m/s.$

The acceleration is defined as the rate of change of velocity, therefore,

$a=\dfrac{\mathrm{d} v}{\mathrm{d}t} \\\Rightarrow v(t)=\int a\ dt = \int (4t-30)\ dt = 4\dfrac{t^2}{2} -30t+c=2t^2-30t+C.$

where, C is the constant of integration,

Since, $v(t) = v_o = 3\ m/s \ \text{at t=0\ s},$

$v_0=2\times 0^2-30\times 0+C=3\\\Rightarrow C=3\ m/s.$

Putting this value of C,

$v(t) = (2t^2-30t+3)\ m/s.$

The velocity is defined as the rate of change of the position, therefore,

$v(t) = \dfrac{\mathrm{d} x}{\mathrm{d}t}\\\Rightarrow x(t) = \int v(t)\ dt=\int (2t^2-30t+3)\ dt =2\dfrac{t^3}3-30\dfrac{t^2}{2}+3t+D.$

where, D is the constant of integration.

Since, $x(t) = x_o = -5\ m/s \ \text{at t=0\ s},$

$x_o=2\dfrac{0^3}3-30\dfrac{0^2}{2}+3\times 0+D=-5\\\Rightarrow D=-5\ m.$

Putting this value of D,

$x(t)=\left (2\dfrac{t^3}3-30\dfrac{t^2}{2}+3t-5 \right)=\left (\dfrac{2t^3}3-15t^2+3t-5\right )\ m.$