To determine how many times the maximum deceleration should be increased to stop a car traveling at 60 m/s in the same shortest distance as when traveling at 20 m/s, we can use the principles of physics related to motion.
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Equations of motion: The stopping distance \( s \) can be related to the initial speed \( u \) and the deceleration \( a \) using the equation: \[ v^2 = u^2 + 2as \] where \( v \) is the final speed (0 m/s when coming to a stop).
Rearranging gives: \[ s = \frac{u^2}{2a} \]
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Initial conditions:
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For the first case (initial speed \( u_1 = 20 , \text{m/s} \), stopping distance \( s = 20 , \text{m} \)): \[ 20 = \frac{(20)^2}{2a_1} \] Rearranging gives: \[ a_1 = \frac{(20)^2}{2 \times 20} = \frac{400}{40} = 10 , \text{m/s}^2 \]
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For the second case (initial speed \( u_2 = 60 , \text{m/s} \)): We want to stop in the same distance \( s = 20 , \text{m} \): \[ 20 = \frac{(60)^2}{2a_2} \] Rearranging gives: \[ a_2 = \frac{(60)^2}{2 \times 20} = \frac{3600}{40} = 90 , \text{m/s}^2 \]
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Comparing deceleration: Now we can find how many times greater \( a_2 \) is compared to \( a_1 \): \[ \text{Factor} = \frac{a_2}{a_1} = \frac{90}{10} = 9 \]
Thus, the maximum deceleration should be increased by a factor of \( 9 \) to stop the car traveling at \( 60 , \text{m/s} \) in the same distance of \( 20 , \text{m} \).
The answer is 4. ×9.
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