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Sagot :
Answer:
Explanation:
Let's confirm the conclusion with the calculation again:
**Car 1:**
- Initial velocity, \( u_1 = 0 \) km/h
- Final velocity, \( v_1 = 0 \) km/h
- Time taken, \( t_1 = 5 \) seconds
Since \( v_1 = u_1 + at_1 \):
\[ 0 = 0 + a \cdot 5 \]
\[ a = 0 \]
Distance traveled by Car 1, \( s_1 \):
\[ s_1 = u_1 t_1 + \frac{1}{2} a t_1^2 \]
\[ s_1 = 0 \cdot 5 + \frac{1}{2} \cdot 0 \cdot 5^2 \]
\[ s_1 = 0 \]
So, Car 1 travels \( s_1 = 0 \) meters after applying the brakes.
**Car 2:**
- Initial velocity, \( u_2 = 3 \) km/h \( = \frac{3}{3.6} \) m/s
- Final velocity, \( v_2 = 0 \) km/h
- Time taken, \( t_2 = 10 \) seconds
Acceleration \( a_2 \):
\[ v_2 = u_2 + a_2 t_2 \]
\[ 0 = \frac{3}{3.6} + a_2 \cdot 10 \]
\[ a_2 = -\frac{3}{3.6 \cdot 10} \]
\[ a_2 = -\frac{1}{12} \]
Distance traveled by Car 2, \( s_2 \):
\[ s_2 = u_2 t_2 + \frac{1}{2} a_2 t_2^2 \]
\[ s_2 = \frac{3}{3.6} \cdot 10 + \frac{1}{2} \cdot \left(-\frac{1}{12}\right) \cdot 10^2 \]
\[ s_2 = \frac{30}{3.6} - \frac{50}{24} \]
\[ s_2 = 8.33 - 2.08 \]
\[ s_2 \approx 6.25 \text{ meters} \]
Therefore, Car 2 travels \( s_2 \approx 6.25 \) meters after applying the brakes.
**Conclusion:**
Car 2 travels farther after the brakes were applied. The distance traveled by Car 2 is \( \boxed{6.25} \) meters, while Car 1 travels \( 0 \) meters. This confirms that Car 2, despite starting at a lower speed initially, travels a greater distance during the braking process due to its longer braking time.
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