s = 12t - 3t2The above equation expresses the distance, s, in meters, traveled by a vehicle in t seconds after the application of brakes. The vehicle which was moving with a velocity of 12 meters/second traveled 12 meters after the application of brakes before coming to a stop. What will be the time taken (in seconds) by the vehicle to stop after the application of brakes?
Question
s = 12t - 3t2The above equation expresses the distance, s, in meters, traveled by a vehicle in t seconds after the application of brakes. The vehicle which was moving with a velocity of 12 meters/second traveled 12 meters after the application of brakes before coming to a stop. What will be the time taken (in seconds) by the vehicle to stop after the application of brakes?
Solution
The given equation is s = 12t - 3t^2. We know that the vehicle comes to a stop after travelling 12 meters. So, we can set s = 12 and solve for t.
12 = 12t - 3t^2
Rearranging the equation gives:
3t^2 - 12t + 12 = 0
This is a quadratic equation in the form of at^2 + bt + c = 0, where a = 3, b = -12, and c = 12. We can solve for t using the quadratic formula, which is t = [-b ± sqrt(b^2 - 4ac)] / 2a.
Substituting the values of a, b, and c into the formula gives:
t = [12 ± sqrt((-12)^2 - 4312)] / 2*3 t = [12 ± sqrt(144 - 144)] / 6 t = [12 ± 0] / 6
So, the possible values for t are t = 12/6 = 2 seconds and t = 0 seconds.
However, t = 0 seconds corresponds to the time at which the brakes were applied. Therefore, the time taken by the vehicle to stop after the application of brakes is 2 seconds.
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