Suppose that the position of a particle is given by [tex]s=f(t)=2 t^3+3 t+9[/tex].

(a) Find the velocity at time [tex]t[/tex].
[tex]v(t)=\square \frac{m}{s}[/tex]

(b) Find the velocity at time [tex]t=3[/tex] seconds.
[tex]\square \frac{m}{s}[/tex]

(c) Find the acceleration at time [tex]t[/tex].
[tex]a(t)=\square \frac{m}{s^2}[/tex]

(d) Find the acceleration at time [tex]t=3[/tex] seconds.
[tex]\square \frac{m}{s^2}[/tex]



Answer :

To analyze the motion of a particle whose position is given by the function [tex]\( s = f(t) = 2t^3 + 3t + 9 \)[/tex], we can find the velocity and acceleration as functions of time, as well as their values at specific instances. Let's go through each part step-by-step.

### Part (a): Find the velocity at time [tex]\( t \)[/tex]

The velocity of the particle is the derivative of the position with respect to time. So we need to differentiate [tex]\( s(t) = 2t^3 + 3t + 9 \)[/tex] with respect to [tex]\( t \)[/tex].

[tex]\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(2t^3 + 3t + 9) \][/tex]

Applying the power rule of differentiation:

[tex]\[ v(t) = 6t^2 + 3 \][/tex]

Thus, the velocity at time [tex]\( t \)[/tex] is:

[tex]\[ v(t) = 6t^2 + 3 \frac{m}{s} \][/tex]

### Part (b): Find the velocity at time [tex]\( t = 3 \)[/tex] seconds

Next, we substitute [tex]\( t = 3 \)[/tex] into the velocity function to find the specific velocity at 3 seconds.

[tex]\[ v(3) = 6(3)^2 + 3 \][/tex]
[tex]\[ v(3) = 6 \cdot 9 + 3 \][/tex]
[tex]\[ v(3) = 54 + 3 \][/tex]
[tex]\[ v(3) = 57 \][/tex]

So, the velocity at [tex]\( t = 3 \)[/tex] seconds is:

[tex]\[ 57 \frac{m}{s} \][/tex]

### Part (c): Find the acceleration at time [tex]\( t \)[/tex]

The acceleration of the particle is the derivative of the velocity with respect to time. So we need to differentiate [tex]\( v(t) = 6t^2 + 3 \)[/tex] with respect to [tex]\( t \)[/tex].

[tex]\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t^2 + 3) \][/tex]

Applying the power rule of differentiation:

[tex]\[ a(t) = 12t \][/tex]

Thus, the acceleration at time [tex]\( t \)[/tex] is:

[tex]\[ a(t) = 12t \frac{m}{s^2} \][/tex]

### Part (d): Find the acceleration at time [tex]\( t = 3 \)[/tex] seconds

Next, we substitute [tex]\( t = 3 \)[/tex] into the acceleration function to find the specific acceleration at 3 seconds.

[tex]\[ a(3) = 12 \cdot 3 \][/tex]
[tex]\[ a(3) = 36 \][/tex]

So, the acceleration at [tex]\( t = 3 \)[/tex] seconds is:

[tex]\[ 36 \frac{m}{s^2} \][/tex]

In summary, we have calculated:
(a) The velocity function [tex]\( v(t) = 6t^2 + 3 \frac{m}{s} \)[/tex]
(b) The velocity at [tex]\( t = 3 \)[/tex] seconds is [tex]\( 57 \frac{m}{s} \)[/tex]
(c) The acceleration function [tex]\( a(t) = 12t \frac{m}{s^2} \)[/tex]
(d) The acceleration at [tex]\( t = 3 \)[/tex] seconds is [tex]\( 36 \frac{m}{s^2} \)[/tex]