Solution 5: Uniform Retardation
Analysis:
Let the distance between Mark-A and Mark-B be $d$. Since marks are at equal distances, the distance between Mark-B and Mark-C is also $d$.
Total distance $AC = 2d$.
Given velocities:
- At A: $v_A = 160$ km/h
- At C: $v_C = 40$ km/h
Step 1: Determine the acceleration (retardation) constant.
Using the equation of motion $v^2 – u^2 = 2as$ for the interval A to C:
$$v_C^2 – v_A^2 = 2a(2d)$$
$$40^2 – 160^2 = 4ad$$
$$1600 – 25600 = 4ad$$
$$-24000 = 4ad \implies ad = -6000$$
Step 2: Locate the point where velocity is 100 km/h.
Let the car travel a distance $x$ from point A to reach 100 km/h. Using the same equation:
$$100^2 – 160^2 = 2ax$$
$$10000 – 25600 = 2ax$$
$$-15600 = 2ax$$
$$ax = -7800$$
Step 3: Compare $x$ with $d$.
We take the ratio of the two results:
$$\frac{ax}{ad} = \frac{-7800}{-6000}$$
$$\frac{x}{d} = 1.3$$
$$x = 1.3d$$
Since $x > d$ (Mark-B is at distance $d$) and $x < 2d$ (Mark-C is at distance $2d$), the point lies between Mark-B and Mark-C.
Answer: (c) Between mark-B and mark-C