A car moving with a speed of 50 km/h can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/h, the minimum stopping distance is 24 m.
Question
A car moving with a speed of 50 km/h can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/h, the minimum stopping distance is 24 m.
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Answer
To solve this problem, we'll need to apply the principles of motion, specifically the third equation of motion which is:
$$ v^2 = u^2 + 2as $$
where:
 ( v ) is the final velocity,
 ( u ) is the initial velocity,
 ( a ) is the acceleration,
 ( s ) is the stopping distance.
StepbyStep :

Convert the Initial Speed:
 For the first part, the car's speed is (50 \ \text{km/h}): $$ 50 \ \text{km/h} = 50 \times \frac{5}{18} = \frac{250}{18} \ \text{m/s} \approx 13.89 \ \text{m/s} $$
 For the second part, the car's speed is (100 \ \text{km/h}): $$ 100 \ \text{km/h} = 100 \times \frac{5}{18} = \frac{500}{18} \ \text{m/s} \approx 27.78 \ \text{m/s} $$

Determine the Deceleration for the Initial Conditions:
 Given:
 ( u = \frac{250}{18} \ \text{m/s} )
 ( v = 0 \ \text{m/s} )
 ( s = 6 \ \text{m} )
 Using the third equation of motion: $$ 0 = \left( \frac{250}{18} \right)^2 + 2 \times a \times 6 $$
 Solving for ( a ): $$ 0 = \frac{62500}{324} + 12a \ 12a =  \frac{62500}{324} \ a = \frac{62500}{324 \times 12} \approx 16.07 \ \text{m/s}^2 $$
 Given:

Calculate the Minimum Stopping Distance for the Increased Speed:
 Given:
 ( u = \frac{500}{18} \ \text{m/s} )
 ( v = 0 \ \text{m/s} )
 ( a = 16.07 \ \text{m/s}^2 )
 Using the third equation of motion again: $$ 0 = \left( \frac{500}{18} \right)^2 + 2 \times (16.07) \times s $$
 Solving for ( s ): $$ 0 = \frac{250000}{324}  32.14s \ 32.14s = \frac{250000}{324} \ s = \frac{250000}{324 \times 32.14} \approx 24 \ \text{m} $$
 Given:
So, the minimum stopping distance when the car is moving at 100 km/h is 24 meters.
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