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1. (a) A particle is projected vertically upwards from the point $p$ - Leaving Cert Applied Maths - Question 1 - 2009
Question 1
1. (a) A particle is projected vertically upwards from the point $p$. At the same instant a second particle is let fall vertically from $q$. The particles meet at $r... show full transcript
Worked Solution & Example Answer:1. (a) A particle is projected vertically upwards from the point $p$ - Leaving Cert Applied Maths - Question 1 - 2009
Step 1
Prove that $|pr| = 3|qr|$
Answer
We start by analyzing the motion of both particles:
-
Particle from point : The initial velocity is denoted as , and the displacement after 2 seconds is given by the equation:
s_p = ut + rac{1}{2} a t^2
For upward motion (with ):
s_p = u(2) - rac{1}{2} g (2^2) = 2u - 2g
Thus, at point : -
Particle from point : The particle falls freely from rest, so using:
s_q = rac{1}{2} g t^2
We find:
qr = rac{1}{2} g (2^2) = 2g
Now, we know that they meet at point , so:
- Equal Speeds Condition:
Since the speeds are equal when they meet, we can derive their speeds:
For :
ightarrow v_q = 2g$$ For $pr$: $$v_p = u - gt ightarrow v_p = u - 2g$$ Setting $v_q = v_p$ yields: $$2g = u - 2g ightarrow u = 4g$$ 4. **Final Step:** Surveying the total movement from points $p$ to $r$ and $q$ to $r$, we can summarize: From $p$ to $r$: For displacements: $$|pr| = 2u - 2g = 2(4g) - 2g = 8g - 2g = 6g$$ For $q$ to $r$: $$|qr| = 2g$$ Therefore: $$|pr| = 3 |qr|$$Step 2
(i) Draw a speed-time graph for the motion of the train.
Answer
The speed-time graph consists of two segments. The first segment shows uniform acceleration from rest to speed , and the second segment shows uniform deceleration back to rest.
- The graph starts at the origin , increases linearly to the maximum speed at time .
- From , it slopes downwards to for uniform deceleration, creating a right triangle.
The graph can be depicted as:
v |
|\
| \
| \
| \
|____\_____
t_1 t_2
Where is the acceleration time and is the total time taken to decelerate to rest.
Step 3
(ii) If the average speed of the train for the whole journey is $ rac{ar{d}}{ rac{ar{3}}{3}}$, find the value of $f$.
Answer
To find the average speed for the whole journey, we will calculate using the formula for average speed:
-
Average Speed Formula:
ext{Average Speed} = rac{ ext{Total Distance}}{ ext{Total Time}}
Given, the average speed is rac{ar{d}}{ rac{ar{3}}{3}}, we rewrite it as:
ightarrow d = vt_{total}$$ 2. **Total Distance:** - Distance during acceleration = $ rac{1}{2} v t_1$ - Distance during deceleration = $ rac{1}{2} v t_2$ Thus, $$d = rac{1}{2} v t_1 + rac{1}{2} v t_2 = rac{v}{2} (t_1 + t_2)$$ 3. **Total Time:** $$t_{total} = t_1 + t_2 = rac{v}{f}$$ 4. **Final Calculation:** Equating: $$ rac{d}{t_{total}} = rac{d}{ rac{v(3f)}{v}} = rac{d}{3f}$$ From this, solve for $f$: $$f = rac{d}{3 imes ext{average speed}}$$ This leads us to conclude that: $$f = 1$$