1. (a) A particle is projected vertically downwards from the top of a tower with speed u m/s. It takes the particle 4 seconds to reach the bottom of the tower. During the third second of its motion the particle travels 29.9 metres. Find (i) the value of u (ii) the height of the tower. (b) A train accelerates uniformly from rest to a speed v m/s. It continues at this speed for a period of time and then decelerates uniformly to rest. In travelling a total distance d metres the train accelerates through a distance pd metres and decelerates through a distance qd metres, where p < 1 and q < 1. (i) Draw a speed-time graph for the motion of the train. (ii) If the average speed of the train for the whole journey is $ \frac{v}{p+q+b} $, find the value of b.

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1. (a) A particle is projected vertically downwards from the top of a tower with speed u m/s - Leaving Cert Applied Maths - Question 1 - 2007

Question 1

1. (a) A particle is projected vertically downwards from the top of a tower with speed u m/s. It takes the particle 4 seconds to reach the bottom of the tower. Duri... show full transcript

Worked Solution & Example Answer:1. (a) A particle is projected vertically downwards from the top of a tower with speed u m/s - Leaving Cert Applied Maths - Question 1 - 2007

Step 1

(i) the value of u

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Answer

To find the value of ( u ), we use the distance formula for uniformly accelerated motion. In the third second, the distance traveled is given by:

$s = u t + \frac{1}{2} a t^2$

For this particular case, the distance traveled in the third second can also be expressed as:

$s = u(2) + \frac{1}{2} a (2^2) - \left( u(1) + \frac{1}{2} a (1^2) \right)$

Setting ( s = 29.9 ) m gives:

$29.9 = u(1) + \frac{1}{2} a (4) - \left( u + \frac{1}{2} a \right)$

This leads to:

$29.9 = u + 2a - \left( u + \frac{1}{2} a \right) \implies 29.9 = 2a - \frac{1}{2} a \implies 29.9 = 2a - \frac{1}{2} a$

Now, calculating the distance from the top to the bottom:

Total distance during the first 4 seconds: [ h = u(4) + \frac{1}{2} a (4^2) ] Substituting this and solving the two equations simultaneously will yield ( u = 5.4 ) m/s.

Step 2

(ii) the height of the tower

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Answer

To calculate the height of the tower, we use the formula again:

[ h = u(4) + \frac{1}{2} a (4^2) ]

Substituting ( u = 5.4 ) m/s, we find acceleration ( a = 9.8 ) m/s²:

[ h = 5.4(4) + \frac{1}{2} (9.8)(16) ]

Calculating gives:

[ h = 21.6 + 78.4 = 100 \text{ m} ]

Step 3

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

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Answer

For the speed-time graph:

The graph starts at the origin (0,0) and increases linearly to a velocity ( v ) at time ( t_1 ).
It remains constant at speed ( v ) from time ( t_1 ) to ( t_2 ).
It then decreases linearly back to the speed of 0 at time ( t_3 ).

The areas under the graph represent the distances travelled during each phase of the motion.

Step 4

(ii) find the value of b

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Answer

The average speed can be calculated using:

[ \text{Average speed} = \frac{d}{t_1 + t_2 + t_3} ]

From the problem, we know:

[ \text{Average speed} = \frac{d}{pd + (d - pd - qd) + qd} = \frac{d}{d} = 1 ]

Comparing with:

[ \text{Average speed} = \frac{v}{p + q + b} ]

Setting both equations equal gives:

[ 1 = \frac{v}{p + q + b} \implies p + q + b = v ]

Thus, we find that the value of ( b = 1 ).

1. (a) A particle is projected vertically downwards from the top of a tower with speed u m/s - Leaving Cert Applied Maths - Question 1 - 2007

Worked Solution & Example Answer:1. (a) A particle is projected vertically downwards from the top of a tower with speed u m/s - Leaving Cert Applied Maths - Question 1 - 2007

(i) the value of u

(ii) the height of the tower

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

(ii) find the value of b

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