Four points a, b, c and d lie on a straight level road. A car, travelling with uniform retardaion, passes point a with a speed of 30 m/s and passes point b with a speed of 20 m/s. The distance from a to b is 100 m. The car comes to rest at d. Find (i) the uniform retardation of the car (ii) the time taken to travel from a to b (iii) the distance from b to d (iv) the speed of the car at c, where c is the midpoint of [bd].

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Four points a, b, c and d lie on a straight level road - Leaving Cert Applied Maths - Question 1 - 2008

Question 1

Four points a, b, c and d lie on a straight level road. A car, travelling with uniform retardaion, passes point a with a speed of 30 m/s and passes point b with a sp... show full transcript

Worked Solution & Example Answer:Four points a, b, c and d lie on a straight level road - Leaving Cert Applied Maths - Question 1 - 2008

Step 1

(i) the uniform retardation of the car

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Answer

To find the uniform retardation, we can use the equation of motion:

$v^2 = u^2 + 2as$

Where:

Final velocity, $v = 20 \, ext{m/s}$ (speed at point b)
Initial velocity, $u = 30 \, ext{m/s}$ (speed at point a)
Distance, $s = 100 \, ext{m}$ (distance from a to b)

Substituting the values, we get:

400 = 900 + 200a\ -500 = 200a\ a = -2.5 \, ext{m/s}^2$$ Thus, the uniform retardation is $2.5 \, ext{m/s}^2$.

Step 2

(ii) the time taken to travel from a to b

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Answer

We can use the formula:

$v = u + at$

Where:

Final velocity, $v = 20 \, ext{m/s}$
Initial velocity, $u = 30 \, ext{m/s}$
Acceleration, $a = -2.5 \, ext{m/s}^2$ (retardation)

Substituting the values:

2.5t = 30 - 20\ t = \frac{10}{2.5} = 4 \, ext{s}$$

Step 3

(iii) the distance from b to d

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Answer

At point b, the car's speed is 20 m/s, and it comes to rest at d. We again use the equation:

$v^2 = u^2 + 2as$

Where:

Final velocity, $v = 0 \, ext{m/s}$ (comes to rest)
Initial velocity, $u = 20 \, ext{m/s}$
Acceleration, $a = -2.5 \, ext{m/s}^2$

Substituting:

0 = 400 - 5s\ 5s = 400\ s = 80 \, ext{m}$$

Step 4

(iv) the speed of the car at c, where c is the midpoint of [bd]

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Answer

The distance from point b to point d is 80 m, so midpoint c is at 40 m from b. We find the speed at c using:

$v^2 = u^2 + 2as$

Where:

Initial velocity, $u = 20 \, ext{m/s}$
Acceleration, $a = -2.5 \, ext{m/s}^2$
Distance, $s = 40 \, ext{m}$ (from b to c)

Substituting:

v^2 = 400 - 200\ v^2 = 200\ v = \sqrt{200} \, ext{or} \, 14.1 \, ext{m/s}$$

Four points a, b, c and d lie on a straight level road - Leaving Cert Applied Maths - Question 1 - 2008

Worked Solution & Example Answer:Four points a, b, c and d lie on a straight level road - Leaving Cert Applied Maths - Question 1 - 2008

(i) the uniform retardation of the car

(ii) the time taken to travel from a to b

(iii) the distance from b to d

(iv) the speed of the car at c, where c is the midpoint of [bd]

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