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Step-by-Step Solution
Step 1: Identify the Relevant Kinematic Equation
When a vehicle comes to a stop under uniform deceleration, the kinematic equation that links initial speed, distance traveled, and acceleration is:
$$v^2 - u^2 = 2 a\,s,$$
where:
$v$ is the final velocity (in this case, $v = 0$),
$u$ is the initial velocity,
$a$ is the (constant) acceleration (negative for deceleration),
$s$ is the distance traveled during deceleration.
Because the car comes to a stop, the equation simplifies to
$$-u^2 = 2a\,s \quad\text{or}\quad u^2 = -2a\,s.$$
We often write it as $u^2 = 2 a\,s,$ keeping in mind that $a$ is negative for deceleration.
Step 2: Write the Equation for the First Car Speed
For the automobile traveling at $u_1 = 60\text{ km/h}$ and stopping in distance $s_1 = 20\text{ m}$:
$$u_1^2 = 2\,a\,s_1.$$
We consider $a$ as the same uniform deceleration in both cases.
Step 3: Write the Equation for the Second Car Speed
When the speed is doubled, $u_2 = 120\text{ km/h}$, and the stopping distance is $s_2$:
$$u_2^2 = 2\,a\,s_2.$$
We assume the same deceleration $a$ applies.
Step 4: Form the Ratio and Solve for $s_2$
Divide the second equation by the first equation:
$$\frac{u_2^2}{u_1^2} = \frac{2\,a\,s_2}{2\,a\,s_1} \quad\Longrightarrow\quad \frac{u_2^2}{u_1^2} = \frac{s_2}{s_1}.$$
Substitute $u_2 = 120\text{ km/h}$, $u_1 = 60\text{ km/h}$, and $s_1 = 20\text{ m}$:
$$\left(\frac{120}{60}\right)^2 = \frac{s_2}{20}.$$
Thus,
$$\left( 2 \right)^2 = \frac{s_2}{20} \quad\Longrightarrow\quad 4 = \frac{s_2}{20} \quad\Longrightarrow\quad s_2 = 80\text{ m}.$$
Step 5: State the Final Answer
The stopping distance when the car is traveling at 120 km/h is 80 m.
