An object with an acceleration of -6.65m/s squared moves from a rate of 78m/s to 2 m/s. How far will the object have traveled during that interval?​

Approximately .

Explanation:

The acceleration of this object is constant. Hence, the SUVAT equation would apply to the motion of this object. Since the duration of this acceleration is not given, apply the SUVAT equation that does not include time.

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Rearrange this equation to find an expression for displacement, :

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In this question, it is given that the initial velocity is , the velocity after acceleration is , and the acceleration is . Evaluate the expression above to find the displacement of this object:

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dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.