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例一:有物自一足够长斜面匀减速上升,由A向B运动,AB长30米。此物在A点初速率为5米/秒,加速度的大小为0.4米/秒~2,求物体由A至B用多少时间?[评析]物体做匀减速运动,初速度v_0=5米/秒,其方向为沿斜面向上,加速度为0.4米/秒~2,方向沿斜面向下,与v_0反向,若取v_0的方向为正向,则a的方向为负向。AB长30米,由A至B,位移沿斜面向上并为正向。将题目所给条件代入匀变速直线运动公式,v_t=v_0+at,s=v_0t+(1/2)at~2。得30=5t-(1/2)×0.4at~2,解此式得t=15秒和t=10秒,这是因为当位移为30米时,t=10秒
Example 1: Something from a sufficiently long slope is evenly decelerated, moving from A to B, and AB is 30 meters long. The initial velocity of this object is 5 meters/second at point A, and the acceleration is 0.4 meters/second~2. How long does it take for object A to B? [Evaluation] The object performs a uniform deceleration movement, and the initial velocity v_0=5 meters/ In seconds, the direction is along the obliquely upward direction, the acceleration is 0.4m/s~2, the direction is along the obliquely downward direction, and is opposite to v_0, if the direction of taking v_0 is positive, then the direction of a is negative. AB is 30 meters long, from A to B. The displacement is obliquely upward and positive. Substituting the condition given by the title into the uniform linear motion formula, v_t=v_0+at,s=v_0t+(1/2)at~2. Get 30=5t-(1/2)×0.4at~2. Solving this equation yields t=15 seconds and t=10 seconds. This is because when the displacement is 30 meters, t=10 seconds.