Two cars, P and Q, travel with the same constant velocity 15 m s⁻¹ along a straight level road. The front of car P is 24 m behind the rear of car Q. At a given instant both cars decelerate, P at 4 m s⁻² and Q at 5 m s⁻². (i) Find, in terms of t, the distance between the cars t seconds later. (ii) Find the distance between the cars when they are at rest. At a particular instant a car of mass 1200 kg is towing a trailer of mass 450 kg on a level road at a speed of 25 m s⁻¹ when the engine exerts a constant power of 50 kW. Friction and air resistance amount to 930 N on the car and 200 N on the trailer. (i) Find the acceleration of the car at this instant. (ii) Calculate the maximum speed at which the car (without the trailer) could travel up an incline of sin⁻¹(1/10) against the same resistance with the engine working at the same rate.

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Two cars, P and Q, travel with the same constant velocity 15 m s⁻¹ along a straight level road - Leaving Cert Applied Maths - Question 1 - 14

Question 1

Two cars, P and Q, travel with the same constant velocity 15 m s⁻¹ along a straight level road. The front of car P is 24 m behind the rear of car Q. At a given insta... show full transcript

Worked Solution & Example Answer:Two cars, P and Q, travel with the same constant velocity 15 m s⁻¹ along a straight level road - Leaving Cert Applied Maths - Question 1 - 14

Step 1

Find, in terms of t, the distance between the cars t seconds later.

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Answer

To find the distance between the cars after t seconds, we start with the equations of motion:

For car P (decelerating at 4 m/s²):

Initial velocity, $u_P = 15$ m/s
After time t seconds, its position can be calculated using:

$s_P = u_P t + \frac{1}{2} a_P t^2 = 15t - 2t^2$ .

For car Q (decelerating at 5 m/s²):

Initial velocity, $u_Q = 15$ m/s
Its position after t seconds is:

$s_Q = u_Q t + \frac{1}{2} a_Q t^2 = 15t - \frac{5}{2}t^2$ .

The initial distance between the two cars is 24 m, so the distance between them after t seconds is:

$d(t) = s_Q - s_P + 24 = \left( 15t - 2.5t^2 \right) - \left( 15t - 2t^2 \right) + 24 = 0.5t^2 + 24$ .

Step 2

Find the distance between the cars when they are at rest.

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Answer

To find the distance when the cars are at rest, we first need to determine the time taken for each car to come to a stop.

For car P: Using the equation $v^2 = u^2 + 2as$ , set $v = 0$ :

$0 = 15^2 + 2(-4)s_P$ Rearranging gives: $s_P = \frac{225}{8} = 28.125$ m.

For car Q: Similarly, $0 = 15^2 + 2(-5)s_Q$ This gives: $s_Q = \frac{225}{10} = 22.5$ m.

Now, the total distance when they are at rest is the initial distance plus the distances traveled: $d = 24 + s_P - s_Q = 24 + 28.125 - 22.5 = 29.625 m$ . Therefore, the distance between the cars when they are at rest is 29.625 m.

Step 3

Find the acceleration of the car at this instant.

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Answer

To find the acceleration, we use the power equation: $P = TV \Rightarrow T = \frac{P}{v} = \frac{50000}{25} = 2000 \text{ N}.$

The net force acting on the car considering resistance is: $F_{net} = T - F_{friction} = 2000 - 930 = 1070 ext{ N}.$

Now applying Newton's second law, $F = ma$ gives us:

$a = \frac{F_{net}}{m} = \frac{1070}{1200} \approx 0.892 \text{ m/s}^2.$ .

Step 4

Calculate the maximum speed at which the car (without the trailer) could travel up an incline of sin⁻¹(1/10).

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Answer

Using the force equilibrium at maximum speed: $T - F_{resistance} - F_{gravity} = 0,$ where the component of gravitational force acting down the incline is $mg \sin(\theta) = 1200g \sin(\sin^{-1}(\frac{1}{10}))$ , which simplifies to: $F_{gravity} = 1200 \times 10 \times \frac{1}{10} = 1200 ext{ N}.$

Setting up the equation: $T - 930 - 1200 = 0 \Rightarrow T = 2130 ext{ N}.$

Finally, using power again: $P = Tv \Rightarrow v = \frac{P}{T} = \frac{50000}{2130} \approx 23.4 ext{ m/s}.$

Two cars, P and Q, travel with the same constant velocity 15 m s⁻¹ along a straight level road - Leaving Cert Applied Maths - Question 1 - 14

Worked Solution & Example Answer:Two cars, P and Q, travel with the same constant velocity 15 m s⁻¹ along a straight level road - Leaving Cert Applied Maths - Question 1 - 14

Find, in terms of t, the distance between the cars t seconds later.

Find the distance between the cars when they are at rest.

Find the acceleration of the car at this instant.

Calculate the maximum speed at which the car (without the trailer) could travel up an incline of sin⁻¹(1/10).

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