A particle moves from point P towards point O. The speed of the particle, tt seconds after passing point P, can be modelled by the formula v=(k−2t)nv = (k - 2t)^n, where kk and nn are constants. The particle has a speed of 27 m/s when t=3.5t = 3.5 and comes to instantaneous rest 4.8 metres to the right of point O when t=8t = 8.

  1. (a)

    Find an expression for the acceleration of the particle in terms of kk, tt and nn.

  2. (b)

    Find the value of kk and of nn.

  3. (c)

    Using the values of kk and nn found in part (b), find the total distance travelled by the particle in the first 8 seconds.

  4. (d)

    Explain whether the given model can continue to describe the motion of the particle after 8 seconds.

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Worked solution

11(a)

1. Differentiate the velocity function with respect to time to find acceleration

a=dvdt=n(k−2t)n−1⋅ddt(k−2t)a = \frac{dv}{dt} = n(k - 2t)^{n-1} \cdot \frac{d}{dt}(k - 2t)
a=−2n(k−2t)n−1a = -2n(k - 2t)^{n-1}

Answer

a=−2n(k−2t)n−1a = -2n(k - 2t)^{n-1}

11(b)

1. Use the condition for instantaneous rest to find $k$.

The particle comes to instantaneous rest when t=8t = 8, so v=0v = 0.
(k−2(8))n=0(k - 2(8))^n = 0
k−16=0k - 16 = 0
k=16k = 16

2. Use the given speed at $t = 3.5$ to form an equation for $n$.

When t=3.5t = 3.5, v=27v = 27. Substituting k=16k = 16:
27=(16−2(3.5))n27 = (16 - 2(3.5))^n
27=(16−7)n27 = (16 - 7)^n
27=9n27 = 9^n

3. Solve the exponential equation to find $n$.

33=(32)n3^3 = (3^2)^n
33=32n3^3 = 3^{2n}
2n=32n = 3
n=1.5n = 1.5

Answer

k=16k = 16, n=1.5n = 1.5

11(c)

1. State the velocity function using the values of $k$ and $n$ from part (b)

From part (b), k=16k = 16 and n=1.5n = 1.5.
v=(16−2t)1.5v = (16 - 2t)^{1.5}

2. Set up the definite integral for the total distance travelled

Since v≥0v \ge 0 for 0≤t≤80 \le t \le 8, the particle does not change direction.
Distance=∫08(16−2t)1.5 dt\text{Distance} = \int_{0}^{8} (16 - 2t)^{1.5} \, dt

3. Integrate the velocity function

∫(16−2t)1.5 dt=(16−2t)2.52.5×(−2)=−(16−2t)2.55\int (16 - 2t)^{1.5} \, dt = \frac{(16 - 2t)^{2.5}}{2.5 \times (-2)} = -\frac{(16 - 2t)^{2.5}}{5}

4. Evaluate the definite integral

Distance=[−(16−2t)2.55]08\text{Distance} = \left[ -\frac{(16 - 2t)^{2.5}}{5} \right]_{0}^{8}
=(−05)−(−162.55)= \left( -\frac{0}{5} \right) - \left( -\frac{16^{2.5}}{5} \right)
=0−(−10245)=204.8= 0 - \left( -\frac{1024}{5} \right) = 204.8

Answer

204.8 m204.8\text{ m}

11(d)

1. Determine the expression for $v$ using the values of $k$ and $n$ from part (b) and evaluate its validity for $t > 8$.

From part (b), k=16k = 16 and n=1.5n = 1.5, so v=(16−2t)1.5v = (16 - 2t)^{1.5}.
For t>8t > 8, 16−2t<016 - 2t < 0.

2. Conclude whether the model can continue to describe the motion.

Since 16−2t16 - 2t is negative for t>8t > 8, (16−2t)1.5(16 - 2t)^{1.5} involves the square root of a negative number, which is undefined. Therefore, the model cannot continue to describe the motion.

Answer

No, because for t>8t > 8, 16−2t<016 - 2t < 0, which makes v=(16−2t)1.5v = (16 - 2t)^{1.5} undefined.

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