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Given Information:

• A particle $P$ is moving along the $x$-axis, and at time $t = 1$ second (where $t > 0$), the acceleration $a$ m/s² of $P$ is given by: $a = 3t - 4.$

• At $t = 0$, P is at rest.

We need to find:

Part (a): Find the velocity of $P$ when $t = 4$.

Acceleration $a$ is the derivative of velocity $v(t)$, so: $a = \frac{dv}{dt}.$ Thus, the velocity $v(t)$ is the integral of acceleration: $v(t) = \int (3t - 4) \, dt.$

Let's integrate: $v(t) = \frac{3t^2}{2} - 4t + C,$ where $C$ is the constant of integration.

We know that at $t = 0$, $v(0) = 0$ because the particle is at rest at $t = 0$. Substituting $t = 0$ into the velocity equation: $0 = \frac{3(0)^2}{2} - 4(0) + C,$ so $C = 0$.

Thus, the velocity function is: $v(t) = \frac{3t^2}{2} - 4t.$

Now, substitute $t = 4$ to find the velocity at that time: $v(4) = \frac{3(4)^2}{2} - 4(4) = \frac{3(16)}{2} - 16 = 24 - 16 = 8 \, \text{m/s}.$

Thus, the velocity when $t = 4$ is: $v(4) = 8 \, \text{m/s}.$

Part (b): At time $T$ seconds, $P$ is instantaneously at rest. Find the value of $T$.

At $t = T$, the velocity is 0: $v(T) = 0.$

We already know the velocity function is: $v(t) = \frac{3t^2}{2} - 4t.$

Substitute $t = T$ into the velocity equation: $0 = \frac{3T^2}{2} - 4T.$

Multiply through by 2 to eliminate the fraction: $0 = 3T^2 - 8T.$

Factor the equation: $0 = T(3T - 8).$

Thus, $T = 0$ or $T = \frac{8}{3}$.

Since $T > 0$, we have: $T = \frac{8}{3} \, \text{seconds}.$

Thus, the time when the particle is instantaneously at rest is: $T = \frac{8}{3} \, \text{seconds}.$

Part (c): Find the displacement of $P$ from the origin when $t = 3$.

The displacement $x(t)$ is the integral of the velocity function: $x(t) = \int v(t) \, dt.$

Substitute $v(t) = \frac{3t^2}{2} - 4t$: $x(t) = \int \left( \frac{3t^2}{2} - 4t \right) \, dt = \frac{3t^3}{6} - 2t^2 + D.$

Simplify: $x(t) = \frac{t^3}{2} - 2t^2 + D.$

To find $D$, we use the initial condition: at $t = 0$, the particle is at the origin, so $x(0) = 0$: $0 = \frac{(0)^3}{2} - 2(0)^2 + D,$ so $D = 0$.

Thus, the displacement function is: $x(t) = \frac{t^3}{2} - 2t^2.$

Now, substitute $t = 3$ to find the displacement at that time: $x(3) = \frac{(3)^3}{2} - 2(3)^2 = \frac{27}{2} - 18 = 13.5 - 18 = -4.5 \, \text{units}.$

Thus, the displacement of $P$ from the origin when $t = 3$ is: $x(3) = -4.5 \, \text{units}.$

Final Answers:

• (a) The velocity when $t = 4$ is $8 \, \text{m/s}$.
• (b) The value of $T$ is $\frac{8}{3} \, \text{seconds}$.
• (c) The displacement when $t = 3$ is $-4.5$ units.

Let me know if you'd like to clarify any steps or need further help!