When an object starts to move in a straight line with a constant acceleration of 0.5m/s ^ 2 from rest, and then decelerates for 6m to rest, if the time taken in the whole process of the object's motion is 10s, the following formula can be obtained: (1) The maximum speed of an object (2) The acceleration of an object during deceleration
Let the acceleration time be T1, and the acceleration be A1, then a1t1 ^ 2 / 2 + 6 = V (average) * t0.5t1 ^ 2 / 2 + 6 = 0.5t1 * 10 / 20.5t1 ^ 2 / 2-2.5t1 + 6 = 0t1 ^ 2-10t1 + 24 = 0, then T1 '= 6S, T1' '= 4S, the maximum velocity VM' = a1t1 '= 3m / s, VM' '= 2m / S (1) the maximum velocity may be 3m / s, or 2m / s
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