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v⃗(12)=2594i^+96j^+82944k^ m/s\vec{v}(12) = 2594\hat{i} + 96\hat{j} + 82944\hat{k} \text{ m/s}

a⃗(12)=432i^+8j^+20736k^ m/s2\vec{a}(12) = 432\hat{i} + 8\hat{j} + 20736\hat{k} \text{ m/s}^2

792i^+16j^+34944k^ m/s792\hat{i} + 16\hat{j} + 34944\hat{k} \text{ m/s}

Step-by-step solution

6 steps

Here's the solution to your first question:

  1. Simplify the position vector.

    First, combine the terms in the position vector:

    r⃗(t)=(6t3+2t)i^+(t2+3t2)j^+12t4k^\vec{r}(t) = (6t^3 + 2t)\hat{i} + (t^2 + 3t^2)\hat{j} + 12t^4\hat{k} r⃗(t)=(6t3+2t)i^+(4t2)j^+12t4k^\vec{r}(t) = (6t^3 + 2t)\hat{i} + (4t^2)\hat{j} + 12t^4\hat{k}
  2. Find the velocity vector (v⃗(t)\vec{v}(t)).

    Velocity is the first derivative of the position vector with respect to time (tt):

    v⃗(t)=dr⃗dt\vec{v}(t) = \frac{d\vec{r}}{dt} v⃗(t)=ddt(6t3+2t)i^+ddt(4t2)j^+ddt(12t4)k^\vec{v}(t) = \frac{d}{dt}(6t^3 + 2t)\hat{i} + \frac{d}{dt}(4t^2)\hat{j} + \frac{d}{dt}(12t^4)\hat{k} v⃗(t)=(18t2+2)i^+(8t)j^+(48t3)k^\vec{v}(t) = (18t^2 + 2)\hat{i} + (8t)\hat{j} + (48t^3)\hat{k}
  3. Find the acceleration vector (a⃗(t)\vec{a}(t)).

    Acceleration is the first derivative of the velocity vector with respect to time (tt):

    a⃗(t)=dv⃗dt\vec{a}(t) = \frac{d\vec{v}}{dt} a⃗(t)=ddt(18t2+2)i^+ddt(8t)j^+ddt(48t3)k^\vec{a}(t) = \frac{d}{dt}(18t^2 + 2)\hat{i} + \frac{d}{dt}(8t)\hat{j} + \frac{d}{dt}(48t^3)\hat{k} a⃗(t)=(36t)i^+(8)j^+(144t2)k^\vec{a}(t) = (36t)\hat{i} + (8)\hat{j} + (144t^2)\hat{k}
  4. Calculate velocity and acceleration at t=12t = 12 sec.

    Substitute t=12t = 12 into the velocity vector:

    v⃗(12)=(18(12)2+2)i^+(8(12))j^+(48(12)3)k^\vec{v}(12) = (18(12)^2 + 2)\hat{i} + (8(12))\hat{j} + (48(12)^3)\hat{k} v⃗(12)=(18(144)+2)i^+(96)j^+(48(1728))k^\vec{v}(12) = (18(144) + 2)\hat{i} + (96)\hat{j} + (48(1728))\hat{k} v⃗(12)=(2592+2)i^+96j^+82944k^\vec{v}(12) = (2592 + 2)\hat{i} + 96\hat{j} + 82944\hat{k} v⃗(12)=2594i^+96j^+82944k^ m/s\vec{v}(12) = \boxed{2594\hat{i} + 96\hat{j} + 82944\hat{k} \text{ m/s}}

    Substitute t=12t = 12 into the acceleration vector:

    a⃗(12)=(36(12))i^+(8)j^+(144(12)2)k^\vec{a}(12) = (36(12))\hat{i} + (8)\hat{j} + (144(12)^2)\hat{k} a⃗(12)=(432)i^+8j^+(144(144))k^\vec{a}(12) = (432)\hat{i} + 8\hat{j} + (144(144))\hat{k} a⃗(12)=432i^+8j^+20736k^ m/s2\vec{a}(12) = \boxed{432\hat{i} + 8\hat{j} + 20736\hat{k} \text{ m/s}^2}
  5. Calculate velocity at t=10t = 10 sec.

    Substitute t=10t = 10 into the velocity vector:

    v⃗(10)=(18(10)2+2)i^+(8(10))j^+(48(10)3)k^\vec{v}(10) = (18(10)^2 + 2)\hat{i} + (8(10))\hat{j} + (48(10)^3)\hat{k} v⃗(10)=(18(100)+2)i^+(80)j^+(48(1000))k^\vec{v}(10) = (18(100) + 2)\hat{i} + (80)\hat{j} + (48(1000))\hat{k} v⃗(10)=(1800+2)i^+80j^+48000k^\vec{v}(10) = (1800 + 2)\hat{i} + 80\hat{j} + 48000\hat{k} v⃗(10)=1802i^+80j^+48000k^ m/s\vec{v}(10) = 1802\hat{i} + 80\hat{j} + 48000\hat{k} \text{ m/s}
  6. Find the difference in velocity between t=12t = 12 and t=10t = 10.

    Difference = v⃗(12)−v⃗(10)\vec{v}(12) - \vec{v}(10)

    Difference=(2594i^+96j^+82944k^)−(1802i^+80j^+48000k^)\text{Difference} = (2594\hat{i} + 96\hat{j} + 82944\hat{k}) - (1802\hat{i} + 80\hat{j} + 48000\hat{k}) Difference=(2594−1802)i^+(96−80)j^+(82944−48000)k^\text{Difference} = (2594 - 1802)\hat{i} + (96 - 80)\hat{j} + (82944 - 48000)\hat{k} Difference=792i^+16j^+34944k^ m/s\text{Difference} = \boxed{792\hat{i} + 16\hat{j} + 34944\hat{k} \text{ m/s}}
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