(a) A particle moving in a straight line experiences a retardation of 0.7v³ m s⁻², where v m s⁻¹ is its speed. It takes 0.04 seconds to reduce its speed from an initial value of 200 m s⁻¹ to v₁, m s⁻¹. Find (i) the value of v₁ (ii) the distance travelled during this 0.04 seconds. (b) A particle moves in a straight line with an acceleration of (2t - 3) m s⁻² at time t seconds. At time t = 0 the particle has velocity of 2 m s⁻¹ and displacement of 1 m relative to a fixed point O on the line. Find (i) the times when the particle changes direction (ii) an expression for the displacement of the particle from O at time t (iii) the total distance travelled in the first 2 seconds.

Leaving Cert Applied Maths Differential Equations: (a) A particle moving in a strai

(a) A particle moving in a straight line experiences a retardation of 0.7v³ m s⁻², where v m s⁻¹ is its speed - Leaving Cert Applied Maths - Question 10 - 2014

Question 10

(a) A particle moving in a straight line experiences a retardation of 0.7v³ m s⁻², where v m s⁻¹ is its speed. It takes 0.04 seconds to reduce its speed from an init... show full transcript

Worked Solution & Example Answer:(a) A particle moving in a straight line experiences a retardation of 0.7v³ m s⁻², where v m s⁻¹ is its speed - Leaving Cert Applied Maths - Question 10 - 2014

Step 1

(i) the value of v₁

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Answer

To find the value of v₁, we start by applying the given retardation formula:

rac{dv}{dt} = -0.7v^3

Integrating both sides over the time interval from 0 to 0.04 seconds,

ext{Let } v(0) = 200 ext{ m/s} ext{ and } v(0.04) = v_1

The integral becomes:

- rac{1}{2v^2} igg|_{200}^{v_1} = -0.7 imes 0.04

Calculating the right side, we have:

- rac{1}{2v_1^2} + rac{1}{2 imes 200^2} = -0.028

Solving for v₁, we find:

rac{1}{2v_1^2} = rac{1}{40000} - 0.028

This leads to:

rac{1}{2v_1^2} = rac{0.028 imes 40000 - 1}{40000} = 0.7 ext{ m/s} herefore v_1 = 4.225 ext{ m/s}

Step 2

(ii) the distance travelled during this 0.04 seconds.

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Answer

To find the distance travelled, we use:

v = v_1 + rac{dv}{ds}

Substituting the values gives us:

Integrating,

- rac{1}{v^2} igg|_{4.225}^{200} = -0.7 imes 0.04

This results in:

rac{1}{4.225^2} - rac{1}{200^2} = 0.028

Solving this results in the distance covered:

The distance travelled over 0.04 seconds is:

$s = 0.33 ext{ m}$

Step 3

(i) the times when the particle changes direction

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Answer

We need to find when the velocity changes from positive to negative, meaning:

rac{dv}{dt} = 2t - 3 = 0

By solving for t:

2t - 3 = 0 ightarrow t = rac{3}{2} ext{ seconds}$$

Step 4

(ii) an expression for the displacement of the particle from O at time t

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Answer

The displacement is found from the velocity:

$rac{ds}{dt} = v = rac{1}{2}(2t^2 - 3t)$

By integrating:

s = rac{1}{2}(t^3 - (1.5)t^{2}) + C\text{ with } C = 1$$

Step 5

(iii) the total distance travelled in the first 2 seconds.

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Answer

Calculating the total distance involves evaluating:

ext{For } t = 1, t = 2\ s = t^3 - rac{3}{2}t^2 + 2t +1\ x = 1 ext{ at } (t=0)

Thus integrating gives us total distance for t from 0 to 2:

$x = rac{11}{6} + (2t +1)\ = 1 m$

(a) A particle moving in a straight line experiences a retardation of 0.7v³ m s⁻², where v m s⁻¹ is its speed - Leaving Cert Applied Maths - Question 10 - 2014

Worked Solution & Example Answer:(a) A particle moving in a straight line experiences a retardation of 0.7v³ m s⁻², where v m s⁻¹ is its speed - Leaving Cert Applied Maths - Question 10 - 2014

(i) the value of v₁

(ii) the distance travelled during this 0.04 seconds.

(i) the times when the particle changes direction

(ii) an expression for the displacement of the particle from O at time t

(iii) the total distance travelled in the first 2 seconds.

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