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The points P and Q lie on a straight level road - Leaving Cert Applied Maths - Question 1 - 2011
Question 1
The points P and Q lie on a straight level road. A car passes P with a speed of 10 ms⁻¹ and accelerates uniformly for 6 seconds to a speed of 22 ms⁻¹. The car then d... show full transcript
Worked Solution & Example Answer:The points P and Q lie on a straight level road - Leaving Cert Applied Maths - Question 1 - 2011
Step 1
Find (i) the acceleration
Answer
To find the acceleration, we use the equation of motion: v = u + at where: v = final velocity = 22 m/s u = initial velocity = 10 m/s t = time = 6 s
Substituting the values:
22 = 10 + a(6)
Rearranging gives: a = \frac{22 - 10}{6} = \frac{12}{6} = 2 , \text{ms}^{-2}.
Step 2
Find (ii) the deceleration
Answer
Using the formula for deceleration: v^2 = u^2 + 2as where: v = final velocity = 18 m/s u = initial velocity = 22 m/s s = distance = 80 m
Substituting the values:
(18)^2 = (22)^2 + 2a(80)
Calculating gives: 324 = 484 + 160a
Rearranging gives: a = \frac{324 - 484}{160} = \frac{-160}{160} = -1 , \text{ms}^{-2}.
Step 3
Find (iii) |PQ|, the distance from P to Q
Answer
The distance from P to Q can be broken into three parts:
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Distance during acceleration: s₁ = ut + \frac{1}{2}at² = 10(6) + \frac{1}{2}(2)(6^2) = 60 + 36 = 96 m.
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Distance during deceleration: 80 m (given).
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Distance during constant speed: s₂ = vt = 18(3) = 54 m.
Therefore, total distance |PQ| = 96 + 80 + 54 = 230 m.
Step 4
Find (iv) the average speed of the car, correct to one decimal place, as it moves from P to Q.
Answer
To find the average speed, we use the formula: \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}
Total distance = 230 m.
Total time = time during acceleration + time during deceleration + time at constant speed:
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Time during acceleration: t₁ = \frac{v - u}{a} = \frac{22 - 10}{2} = 6 , \text{s}.
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Time during deceleration: t₂ = \frac{v - u}{a} = \frac{18 - 22}{-1} = 4 , \text{s}.
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Time at constant speed: t₃ = \frac{s}{v} = \frac{54}{18} = 3 , \text{s}.
Total time = 6 + 4 + 3 = 13 s.
Thus, average speed = \frac{230}{13} \approx 17.7 , \text{ms}^{-1}.