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Step-by-Step Solution
Step 1: Identify the Known Quantities (Case 1)
Initially, the car travels at a speed of 50 km/hr and covers a minimum stopping distance of 6 m. Let:
Initial speed (u) = 50 km/hr
Final speed (v) = 0 (car comes to rest)
Displacement (s) = 6 m
Deceleration (a) = ? (to be found)
Step 2: Convert the Speed from km/hr to m/s (Case 1)
To convert 50 km/hr to m/s, use the conversion factor
50 km/hr =
$ \frac{50 \times 1000}{3600} $ m/s.
Thus,
$ u = \frac{50 \times 1000}{3600} = \frac{125}{9} \text{ m/s}. $
Step 3: Apply the Kinematic Equation (Case 1)
Use the equation
$ v^{2} = u^{2} + 2 \, a \, s. $
Here, $v = 0$, so the equation becomes:
$ 0 = u^{2} + 2\,a\,s. $
Solving for $a$,
$ a = - \frac{u^{2}}{2s}. $
Substitute $u = \tfrac{125}{9}$ m/s and $s = 6$ m:
$ a = - \frac{\left(\tfrac{125}{9}\right)^{2}}{2 \times 6}\ \text{m/s}^2. $
Numerically, this evaluates to approximately
$ a = -16\ \text{m/s}^2. $
Step 4: Set Up the Known Quantities for the Second Case
For the second scenario, the same car moves at 100 km/hr. We use the same deceleration $a = -16$ m/s2. Let:
Initial speed (u) = 100 km/hr
Final speed (v) = 0
Deceleration (a) = -16 m/s2
Displacement (s) = ? (to be found)
Step 5: Convert the Speed from km/hr to m/s (Case 2)
To convert 100 km/hr to m/s:
$ u = \frac{100 \times 1000}{3600} = \frac{250}{9} \text{ m/s}. $
Step 6: Apply the Kinematic Equation (Case 2)
Again, use
$ v^{2} = u^{2} + 2 \, a \, s. $
Since $v = 0$,
$ 0 = u^{2} + 2\,a\,s \quad \Longrightarrow \quad s = -\frac{u^{2}}{2a}.
$
Substitute $u = \tfrac{250}{9}$ m/s and $a = -16$ m/s2:
$ s = - \frac{\left(\tfrac{250}{9}\right)^{2}}{2 \times (-16)}. $
This calculation gives
$ s = 24\ \text{m}.
$
Final Answer
The minimum stopping distance when the car is moving at 100 km/hr is 24 m.
