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 $T$ in the rope, we analyze the forces acting on the car only (as the tension is responsible for its acceleration).

Applying Newton's second law to the car, we have:

$T - 400 = 800a$

Substituting the value of acceleration $a = 0.7 \, \text{m s}^{-2}$:

$T - 400 = 800 \times 0.7$

Thus:

$T - 400 = 560$

Therefore, solving for $T$:

$T = 560 + 400 = 960 \, \text{N}.$

When the rope breaks, the truck continues with the same driving force, so we analyze the new acceleration $a'$ of the truck:

$F_{net} = F_{driving} - F_{resistance} = 2400 \, \text{N} - 600 \, \text{N} = 1800 \, \text{N}$

The mass of the truck is 1200 kg, so the new acceleration is:

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

Next, we calculate the time taken ($t$) to reach 28 m s⁻¹:

Using the formula:

$v = u + a't$

Substituting $v = 28 \, \text{m s}^{-1}$, $u = 20 \, \text{m s}^{-1}$, and $a' = 1.5 \, \text{m s}^{-2}$:

$28 = 20 + 1.5t$

Thus:

$1.5t = 8 \, \Rightarrow t = \frac{8}{1.5} \approx 5.33 \, ext{s}.$

If the rope had not broken, the previous acceleration is used (0.7 m s²):

$a = 0.7 \, \text{m s}^{-2}$

The time taken to reach the same speed becomes:

Using the same final speed formula:

$t = \frac{28 - 20}{0.7} = \frac{8}{0.7} \approx 11.43 \, ext{s}.$

The difference in time is:

$11.43 - 5.33 \approx 6.1 \, ext{s}.$