1. (a) A ball is thrown vertically upwards with a speed of 44.1 m s⁻¹. Calculate the time interval between the instants that the ball is 39.2 m above the point of projection. (b) A lift ascends from rest with constant acceleration until it reaches a speed v. It continues at this speed for t₁ seconds and then decelerates uniformly to rest with deceleration f. The total distance ascended is d, and the total time taken is s seconds. (i) Draw a speed-time graph for the motion of the lift. (ii) Show that v = 1/2 f (t₁ - t₀). (iii) Show that t₁ = √(4d/f).

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1. (a) A ball is thrown vertically upwards with a speed of 44.1 m s⁻¹ - Leaving Cert Applied Maths - Question 1 - 2013

Question 1

1. (a) A ball is thrown vertically upwards with a speed of 44.1 m s⁻¹. Calculate the time interval between the instants that the ball is 39.2 m above the point of p... show full transcript

Worked Solution & Example Answer:1. (a) A ball is thrown vertically upwards with a speed of 44.1 m s⁻¹ - Leaving Cert Applied Maths - Question 1 - 2013

Step 1

Calculate the time interval for (a)

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Answer

To find the time interval, we use the kinematic equation:

$s = ut + \frac{1}{2}at^2$

In this case, the initial velocity, $u = 44.1$ m/s, and the displacement, $s = 39.2$ m, and the acceleration $a = -9.8$ m/s² (because gravity is acting downwards).

Substituting these values:

$39.2 = 44.1t - \frac{1}{2}(9.8)t^2$

This simplifies to:

$39.2 = 44.1t - 4.9t^2$

Rearranging gives:

$4.9t^2 - 44.1t + 39.2 = 0$

Using the quadratic formula, $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ :

Where $a = 4.9, b = -44.1, c = 39.2$ :

Calculate the discriminant: $D = b^2 - 4ac = (-44.1)^2 - 4(4.9)(39.2) = 1959.61 - 768.16 = 1191.45$
Finding the roots: $t = \frac{44.1 \pm \sqrt{1191.45}}{2(4.9)}$ This gives two time values, from which we find:
- The relevant time intervals for when the ball is at 39.2 m.

Finally, calculate the time interval as the difference of those two times.

Step 2

Draw a speed-time graph for the motion of the lift (b)(i)

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Answer

The speed-time graph consists of three segments:

Acceleration Phase: The graph starts at (0,0) and rises linearly to (t₁, v) as the lift accelerates.
Constant Speed Phase: It continues horizontally from (t₁, v) to (t₁ + t₀, v).
Deceleration Phase: The graph then slopes downwards back to (t₁ + t₀ + t₂, 0) where the lift comes to rest. This forms a trapezoidal shape on the graph.

Step 3

Show that v = 1/2 f (t₁ - t₀) (b)(ii)

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Answer

Using the definition of average speed:

$v = \frac{d}{t}$

The lift first accelerates, then moves at constant speed for $t₁$ , and then decelerates. The acceleration phase can be described as: $f = \frac{v}{t₁ - t₀}$ Thus, rearranging gives: $v = f(t₁ - t₀)$ Therefore, to show: $v = \frac{1}{2}f(t₁ - t₀)$ Consider the average speed during acceleration, we have: $v = \frac{f}{2}(t₁ - t₀).$

Step 4

Show that t₁ = √(4d/f) (b)(iii)

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Answer

Starting from the kinematic equation:

$d = \frac{1}{2} f t^2$

For constant acceleration, if we rearrange for $t$ :

$t^2 = \frac{2d}{f}$

Thus, taking the square root: $t = \sqrt{\frac{2d}{f}}$ Considering the acceleration time equals $t₁$ , we can write: $t₁ = \sqrt{\frac{4d}{f}}.$

1. (a) A ball is thrown vertically upwards with a speed of 44.1 m s⁻¹ - Leaving Cert Applied Maths - Question 1 - 2013

Worked Solution & Example Answer:1. (a) A ball is thrown vertically upwards with a speed of 44.1 m s⁻¹ - Leaving Cert Applied Maths - Question 1 - 2013

Calculate the time interval for (a)

Draw a speed-time graph for the motion of the lift (b)(i)

Show that v = 1/2 f (t₁ - t₀) (b)(ii)

Show that t₁ = √(4d/f) (b)(iii)

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