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1. (a) A particle is released from rest at A and falls vertically passing two points B and C - Leaving Cert Applied Maths - Question 1 - 2011
Question 1
1. (a) A particle is released from rest at A and falls vertically passing two points B and C. It reaches B after t seconds and takes \( \frac{2}{7} \) seconds to fa... show full transcript
Worked Solution & Example Answer:1. (a) A particle is released from rest at A and falls vertically passing two points B and C - Leaving Cert Applied Maths - Question 1 - 2011
Step 1
A particle is released from rest at A and falls vertically passing two points B and C.
Answer
Let the height between points A and B be ( h_1 ) and the height between points B and C be 2.45 m.
Using the formula for the distance fallen under gravity: [ s = ut + \frac{1}{2}gt^2 ] where ( u = 0 ), we determine:
For AB: [ h_1 = 0 + \frac{1}{2}g(t^2) ]
For AC (total distance): [ h_1 + 2.45 = 0 + \frac{1}{2}g\left( t + \frac{2}{7} \right)^2 ]
Substituting ( g = 9.8 ) m/s² and solving for ( t ) results in: [ t = \frac{7}{8} \text{ seconds} ]
Step 2
A car accelerates uniformly from rest to a speed v in t1 seconds.
Answer
The average speed for the journey can be calculated as follows: [ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{\frac{1}{2}vt_1 + vt + \frac{1}{2}vt_2}{t_1 + t + t_2} ] Given that this average speed equals ( \frac{3v}{4} ), we can set up the following equation: [ 3v(t_1 + t + t_2) = 4\left( \frac{1}{2}vt_1 + vt + \frac{1}{2}vt_2 \right) ]
Solving this, we find: [ t_1 + t_2 = \frac{2}{3}t ]
Step 3
If a speed limit of \( \frac{2v}{3} \) were to be applied, find in terms of t the least time the journey would have taken.
Answer
When the speed limit is applied, the time consumed will be: [ t_1 + t + t_2 = \frac{t}{3} + t + \frac{2}{3}t = 2t ] Thus: [ t_1 + t_2 + t = \frac{4}{3}t ] This indicates that the least time the journey would have taken is: [ t_1 + t_2 + t = \frac{8}{3}t ]