Question

van object moving with uniform acceleration has a velocity of 10.0 cm/s in the positive x-direction when its x-coordinate is 3.09 cm. if its x-coordinate 2.55 s later is −5.00 cm, what is its acceleration?

Answer 1

Initial velocity of object vi=10.0 cm/s

initial position fo vector of the object is xi=3.09 cm

Final position of vector xf=-5.00cm

then displacement of object s = xf-xi=-5.00-3.09=-8.09cm

time t=2.55 s

s=vit+1/2at2

-5.00 = 11*2.55+1/2*a2.552

a = (-5.00 - 10*2.55*2)/2.552 = 2.94 cm

Acceleration is 2.94 cm.

What is acceleration?

Speed increase is the name we provide for any cycle where the speed changes. Since speed is a speed and a bearing, there are simply two different ways for you to speed up: change your speed or shift your course or change both. In mechanics, speed increase is the pace of progress of the speed of an item concerning time. Speed increases are vector amounts. The direction of an item's speed increase is given by the direction of the net power following up on that article. An item's typical speed increase throughout some stretch of time is its adjustment of speed separated by the term of the period. Numerically, quick speed increase, in the meantime, is the constraint of the typical speed increase over a little time period. In the terms of analytics, immediate speed increase is the subordinate of the speed vector concerning time.

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