10. (a) At time / seconds the acceleration a m s² of a particle, P, is given by a = 8t + 4. At t = 0, P passes through a fixed point with velocity -24 m s⁻¹. (i) Show that P changes its direction of motion only once in the subsequent motion. (ii) Find the distance travelled by P between t = 0 and t = 3. (b) A particle moves along a straight line in such a way that its acceleration is always directed towards a fixed point O on the line, and is proportional to its displacement from that point. The displacement of the particle from O at time t is x. The equation of motion is $ \frac{dv}{dx} = -\omega^2 x $ where v is the velocity of the particle at time t and $ \omega $ is a constant. The particle starts from rest at a point P, a distance A from O. Derive an expression for (i) v in terms of A, $ \omega $ and x (ii) x in terms of A, $ \omega $ and t.

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10. (a) At time / seconds the acceleration a m s² of a particle, P, is given by a = 8t + 4 - Leaving Cert Applied Maths - Question 10 - 2016

Question 10

10. (a) At time / seconds the acceleration a m s² of a particle, P, is given by a = 8t + 4. At t = 0, P passes through a fixed point with velocity -24 m s⁻¹. ... show full transcript

Worked Solution & Example Answer:10. (a) At time / seconds the acceleration a m s² of a particle, P, is given by a = 8t + 4 - Leaving Cert Applied Maths - Question 10 - 2016

Step 1

Show that P changes its direction of motion only once in the subsequent motion.

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Answer

To analyze when the particle changes direction, we need to determine the velocity function v(t). The acceleration is given by:

$a = \frac{dv}{dt} = 8t + 4$

Integrating with respect to time, we get:

$v = 8t^2/2 + 4t + C$

We know that at time t = 0, the velocity v = -24 m/s, hence:

$-24 = 0 + 0 + C \Rightarrow C = -24$

Thus, the velocity function becomes:

$v = 4t^2 + 4t - 24$

To find the points at which the direction of motion changes, we set v = 0:

$4t^2 + 4t - 24 = 0 \Rightarrow t^2 + t - 6 = 0$

Factoring the equation:

$(t + 3)(t - 2) = 0 \Rightarrow t = -3 \text{ or } t = 2$

Since time cannot be negative, the particle changes direction at t = 2 seconds.

Step 2

Find the distance travelled by P between t = 0 and t = 3.

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Answer

We need to compute the distance travelled using the position function s(t). We integrate the velocity function:

Since we had:

$v = 4t^2 + 4t - 24$

Integrating, we get:

$s = \int v \, dt = \int (4t^2 + 4t - 24) \, dt = \frac{4}{3}t^3 + 2t^2 - 24t + C_1$

Setting conditions at t = 0:

$s(0) = 0 \Rightarrow C_1 = 0$

Thus, we have:

$s = \frac{4}{3}t^3 + 2t^2 - 24t$

Calculating distance from t = 0 to t = 3:

$s(3) = \frac{4}{3}(3^3) + 2(3^2) - 24(3) \Rightarrow s(3) = 36 - 72 = -36$ .

For t = 0 to t = 2, we calculate:

= \frac{32}{3} + 8 - 48 = \frac{32}{3} + \frac{24}{3} - \frac{144}{3} = -\frac{88}{3} \approx -29.33.$$ Now, total distance travelled is given by the total magnitude while changing direction: Distance from t=0 to t=2: |s(2)| = 29.33, Distance from t=2 to t=3: s(3) - s(2) = |-36 + 29.33| = | -6.67| = 6.67 Thus the total distance is: $$ ext{Total distance} = 29.33 + 6.67 = 36.00 m $$.

Step 3

Derive an expression for v in terms of A, $ \omega $ and x.

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Answer

Starting from the equation of motion:

$\frac{dv}{dx} = -\omega^2 x$

We can integrate with respect to x:

$\int dv = -\omega^2 \int x \, dx \Rightarrow v = -\frac{1}{2}\omega^2 x^2 + C$

Since the particle starts from rest at point P (i.e., v = 0 when x = A):

$0 = -\frac{1}{2}\omega^2 A^2 + C \Rightarrow C = \frac{1}{2}\omega^2 A^2$

Thus,

$v = -\frac{1}{2}\omega^2 x^2 + \frac{1}{2}\omega^2 A^2$

Rearranging:

$v = \frac{\omega^2}{2}(A^2 - x^2)$ .

Step 4

Derive an expression for x in terms of A, $ \omega $ and t.

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Answer

To find x in terms of time, integrate velocity:

Starting from:

$v = -\frac{1}{2}\omega^2 A^2 + \frac{1}{2}\omega^2 x^2$

Rearranging gives:

$\frac{dx}{dt} = \frac{1}{2}\omega^2 A^2 - \frac{1}{2}\omega^2 x^2$

Let's integrate:

$\int \frac{dx}{A^2 - x^2} = -\int \omega^2 dt$

Using the integral formula:

$\sin^{-1} \left( \frac{x}{A} \right) = -\alpha t + C_1$

Thus, we have:

$x = A \sin(-\alpha t + C_1)$

By using initial condition (at t = 0, x = A):

So:

$A = A \sin(C_1) \Rightarrow C_1 = \frac{\pi}{2}$

Final expression becomes:

$x = A \cos(\alpha t)$ .

10. (a) At time / seconds the acceleration a m s² of a particle, P, is given by a = 8t + 4 - Leaving Cert Applied Maths - Question 10 - 2016

Worked Solution & Example Answer:10. (a) At time / seconds the acceleration a m s² of a particle, P, is given by a = 8t + 4 - Leaving Cert Applied Maths - Question 10 - 2016

Show that P changes its direction of motion only once in the subsequent motion.

Find the distance travelled by P between t = 0 and t = 3.

Derive an expression for v in terms of A, \( \omega \) and x.

Derive an expression for x in terms of A, \( \omega \) and t.

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