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3 points p, q and r lie on a straight level road - Leaving Cert Applied Maths - Question 1 - 2009
Question 1
3 points p, q and r lie on a straight level road. Two cars, A and B, are moving towards each other on the road. Car A passes p with speed 3 m/s and uniform acceler... show full transcript
Worked Solution & Example Answer:3 points p, q and r lie on a straight level road - Leaving Cert Applied Maths - Question 1 - 2009
Step 1
Find (i) the speed of car A and the speed of car B at q.
Answer
To calculate the speed of car A and car B at point q after 7 seconds:
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For car A: Using the equation of motion: v = u + at Where:
- Initial speed (u) = 3 m/s
- Acceleration (a) = 2 m/s²
- Time (t) = 7 s
Substituting values: v_A = 3 + (2 × 7) v_A = 3 + 14 = 17 m/s
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For car B: Similarly, using the same formula: v = u + at Where:
- Initial speed (u) = 5 m/s
- Acceleration (a) = 4 m/s²
Substituting values: v_B = 5 + (4 × 7) v_B = 5 + 28 = 33 m/s
Step 2
Find (ii) |pq| and |qr|, the distances A and B have moved in these 7 s.
Answer
To find the distances moved by both cars A and B during the 7 seconds:
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Distance for car A: Using the formula for distance: s = ut + \frac{1}{2}at^2 Where:
- u = 3 m/s, a = 2 m/s², t = 7 s
Substituting values: s_A = (3 × 7) + \frac{1}{2}(2)(7^2) s_A = 21 + \frac{1}{2}(2)(49) s_A = 21 + 49 = 70 m
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Distance for car B: Similarly: s = ut + \frac{1}{2}at^2 Where:
- u = 5 m/s, a = 4 m/s², t = 7 s
Substituting values: s_B = (5 × 7) + \frac{1}{2}(4)(7^2) s_B = 35 + \frac{1}{2}(4)(49) s_B = 35 + 98 = 133 m
Step 3
Find (iii) the total time for car A to travel from p to r.
Answer
To find the total time for car A to travel from p to r:
The distance from p to q |pq| is 70 m and from q to r |qr| is 133 m. Hence, the total distance is:
total distance = |pq| + |qr| = 70 m + 133 m = 203 m
After passing q, car A moves to r at a uniform speed. At q, its final speed:
v_A = 17 m/s
The time taken to travel from q to r: t = \frac{distance}{speed} t = \frac{133}{17} \approx 7.8 s
Now add the time taken to accelerate to 7 seconds:
total time = 7 + 7.8 = 14.8 s