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Step-by-Step Solution
Step 1: Identify the Known Quantities for the First Case
• Initial speed of the automobile in the first case, $u_1 = 40\ \text{km/h}$.
• Final speed, $v_1 = 0\ \text{km/h}$ (the automobile comes to rest).
• Stopping distance in the first case, $s_1 = 40\ \text{m}$.
• We assume uniform deceleration (negative acceleration) $a$.
Step 2: Expressing the Equation of Motion
The equation of motion under constant acceleration (or deceleration) is:
$$v^2 - u^2 = 2as$$
In the first case:
$$0^2 - (u_1)^2 = 2 \, a \, s_1$$
So,
$$-(u_1)^2 = 2\,a\,(40)$$
(1)
Step 3: Setting Up for the Second Case
• Initial speed in the second case, $u_2 = 80\ \text{km/h}$.
• Final speed, $v_2 = 0\ \text{km/h}$ (again the automobile comes to rest).
• Let the stopping distance in the second case be $s_2$.
• The same deceleration $a$ acts.
Using the same equation of motion:
$$0^2 - (u_2)^2 = 2\,a\,s_2$$
So,
$$-(u_2)^2 = 2\,a\,s_2$$
(2)
Step 4: Relating the Two Cases
Divide equation (2) by equation (1):
$$\frac{-(u_2)^2}{-(u_1)^2} = \frac{2\,a\,s_2}{2\,a\,s_1}$$
This simplifies to:
$$\frac{(u_2)^2}{(u_1)^2} = \frac{s_2}{s_1}$$
Since $u_2 = 80\ \text{km/h}$ and $u_1 = 40\ \text{km/h}$,
$$\frac{80^2}{40^2} = \frac{s_2}{40}$$
$$\frac{6400}{1600} = \frac{s_2}{40}$$
$$4 = \frac{s_2}{40} \implies s_2 = 4 \times 40 = 160\ \text{m}$$
Step 5: Conclude the Minimum Stopping Distance
The required stopping distance for the automobile moving at $80\ \text{km/h}$ is:
$$s_2 = 160\ \text{m}$$
