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.

Step-by-Step :

  1. 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}): $$ 100 \ \text{km/h} = 100 \times \frac{5}{18} = \frac{500}{18} \ \text{m/s} \approx 27.78 \ \text{m/s} $$
  2. 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 ): $$ 0 = \frac{62500}{324} + 12a \ -12a = - \frac{62500}{324} \ a = \frac{62500}{324 \times 12} \approx 16.07 \ \text{m/s}^2 $$
  3. 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 ): $$ 0 = \frac{250000}{324} - 32.14s \ 32.14s = \frac{250000}{324} \ s = \frac{250000}{324 \times 32.14} \approx 24 \ \text{m} $$

So, the minimum stopping distance when the car is moving at 100 km/h is 24 meters.


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