A car which has run out of petrol is being towed by a breakdown truck along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 8 - 2003 - Paper 1

To find the tension in the rope, we consider only the car and the forces acting on it. The driving force (tension T) acting on the car is countered by the resistive force:

T - 400 N = 800 kg * a

Substituting the value of acceleration:

T - 400 = 800 * 0.7

Thus, we proceed to solve for T:

T - 400 = 560

This leads to:

T = 560 + 400 = 960 N.

After the rope breaks, the truck continues to move under the engine's driving force, with a net force given by:

$F_{net} = 2400 ext{ N} - 600 ext{ N} = 1800 ext{ N}$

Calculating the new acceleration (a') of the truck:

$a' = \frac{F_{net}}{m} = \frac{1800}{1200} = 1.5 ext{ m s}^{-2}$

Next, we will determine the time it takes for the truck to reach a speed of 28 m s⁻¹:

Using the equation of motion:

$v = u + at$

where v = 28 m s⁻¹, u = 20 m s⁻¹, and a = 1.5 m s⁻¹:

$28 = 20 + 1.5 t$

Solving for t gives:

$t = \frac{28 - 20}{1.5} = 5.33 ext{ s}$

Now, if the rope had not broken, we would use the previously calculated acceleration (0.7 m s⁻¹):

$a = 0.7 ext{ m s}^{-2}$

Time to reach 28 m s⁻¹ under normal circumstances:

Using the same equation:

$28 = 20 + 0.7 t$

Solving for t gives:

$t = \frac{28 - 20}{0.7} = 11.43 ext{ s}$

Finally, the difference in time:

$11.43 - 5.33 = 6.1 ext{ s}$

Thus, we show that the truck reaches a speed of 28 m s⁻¹ approximately 6 s earlier than if the rope had not broken.