A truck of mass 1750 kg is towing a car of mass 750 kg along a straight horizontal road. The two vehicles are joined by a light towbar which is inclined at an angle θ to the road, as shown in Figure 4. The vehicles are travelling at 20 m s⁻¹ at the zone where the speed limit is 14 m s⁻¹. The truck's brakes are applied to give a constant braking force on the truck. The distance travelled between the instant when the brakes are applied and the instant when the speed of each vehicle is 14 m s⁻¹ is 100 m. (a) Find the deceleration of the truck and the car. (b) The constant braking force on the truck has magnitude R newtons. The truck and the car also experience constant resistances to motion of 500 N and 300 N respectively. Given that cos θ = 0.9, find (c) the force in the towbar, (d) the value of R.

A-Level Edexcel Maths Mechanics Newtons Second Law: A truck of mass 1750 kg is towin

A truck of mass 1750 kg is towing a car of mass 750 kg along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 7 - 2013 - Paper 2

Question 7

A truck of mass 1750 kg is towing a car of mass 750 kg along a straight horizontal road. The two vehicles are joined by a light towbar which is inclined at an angle ... show full transcript

Worked Solution & Example Answer:A truck of mass 1750 kg is towing a car of mass 750 kg along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 7 - 2013 - Paper 2

Step 1

Find the deceleration of the truck and the car.

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Answer

To determine the deceleration of both vehicles, we can use the equation of motion:

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

Where:

Final velocity, $v = 14 \, \text{m s}^{-1}$
Initial velocity, $u = 20 \, \text{m s}^{-1}$
Distance, $s = 100 \, \text{m}$
Acceleration, $a$ is what we need to find.

Rearranging gives:

$0 = (20)^2 + 2a(100)$

$2a(100) = -400$

$a = -2 \, \text{m s}^{-2}$

Thus, the deceleration is $2 \, \text{m s}^{-2}$ for both the truck and the car.

Step 2

The constant braking force on the truck has magnitude R newtons. The truck and the car also experience constant resistances to motion of 500 N and 300 N respectively. Given that cos θ = 0.9, find

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Answer

Let’s calculate the horizontal forces acting on the car:

The net force in the horizontal direction for the car can be expressed as:

$R_c = T - 300$

Where T is the tension in the towbar. Setting $R_c = m imes a$ , we get:

$750 \cdot (-2) = T - 300$

So:

$T = 1500 - 300 \Rightarrow T = 1200 \, \text{N}$

Now for the truck, we have:

The horizontal forces acting:

$T \cdot \cos \theta - 500 - R = 1750 \cdot (-2)$

Substituting in the value of T:

$1200 \cdot 0.9 - 500 - R = -3500$

Now calculate:

$1080 - 500 - R = -3500$

This leads to:

$580 - R = -3500\Rightarrow R = 580 + 3500 = 4080 \, \text{N}$

Step 3

the value of R.

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Answer

From the previous calculation, we found that the magnitude of the constant braking force R on the truck is:

$R = 4080 \, \text{N}$

A truck of mass 1750 kg is towing a car of mass 750 kg along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 7 - 2013 - Paper 2

Worked Solution & Example Answer:A truck of mass 1750 kg is towing a car of mass 750 kg along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 7 - 2013 - Paper 2

Find the deceleration of the truck and the car.

The constant braking force on the truck has magnitude R newtons. The truck and the car also experience constant resistances to motion of 500 N and 300 N respectively. Given that cos θ = 0.9, find

the value of R.

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