Answer :
The time it takes the car to stop is 1.34 s.
The initial speed of the car is 77.46 ft/s
The given parameters;
acceleration of the car, a = 24 ft/s²
distance traveled by the car, s = 125 ft
The initial speed of the car is calculated as;
When the car stops, the final velocity will be zero,
[tex]v^2 = u^2 + 2as\\\\0 = u^2 + 2(-24\times 125)\\\\u^2 = 6000\\\\u = \sqrt{6000} \\\\u = 77.46 \ ft/s[/tex]
The time taken for the car to stop is calculated as;
[tex]s = ut + \frac{1}{2} at^2\\\\125 = 77.46t + 0.5\times 24\times t^2\\\\125 = 77.46t + 12t^2\\\\12t^2 + 77.46t - 125 = 0\\\\solve \ the \ quadratic \ equation \ using \ formula \ method;\\\\a = 12, \ b = 77.46, \ c = - 125\\\\t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-77.46 \ \ +/- \ \ \sqrt{(77.46)^2 - 4(12\times -125)} }{2(12)}\\\\t = 1.34 \ s[/tex]
Thus, the time it takes the car to stop is 1.34 s.
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