A car is towing a trailer along a straight horizontal road by means of a horizontal tow-rope - Edexcel - A-Level Maths Mechanics - Question 6 - 2006 - Paper 1

To find the tension in the tow-rope, we will consider the trailer:

Using Newton's second law for the trailer:

• Let T be the tension in the tow-rope:

For the trailer:

$T - 280 = 700a$

Substituting the acceleration:

$T - 280 = 700 \times 0.7$

Calculating:

$T - 280 = 490$

So:

$T = 490 + 280 = 770 N$

After the tow-rope breaks, the car continues to move with an initial velocity of 12 m/s. As there are no additional forces acting on the car in this time interval, the acceleration is 0. The distance can be calculated using:

$s = ut + \frac{1}{2} a t^2$

Where:

• u = initial velocity (12 m/s)
• a = 0 (after the tow-rope breaks)
• t = time (4 s)

Calculating the distance:

$s = 12 \times 4 + \frac{1}{2} \times 0 \times 4^2 = 48 \, \text{m}$

Thus, the car moves 48 m in the first 4 seconds.

The modelling assumption that the tow-rope is inextensible implies that the distance between the car and trailer remains constant while the tow-rope is under tension. This means both the car and trailer will accelerate together at the same rate until the rope breaks. Once the tow-rope breaks, the car continues to move forward independently, which is key to calculating the motion of both vehicles.