A particle is moving with a velocity of 60.0 m/s

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A particle initially moves at a velocity of 60.0 m/s in the positive x direction and uniformly decelerates to zero over a 15.0-second interval. The acceleration can be calculated using the formula a = (final velocity - initial velocity) / time, resulting in an acceleration of -4.0 m/s². The negative sign indicates that the particle is decelerating, which is significant as it shows a change in direction of motion. Properly applying the formula and clearly presenting calculations are emphasized for accuracy. Understanding the implications of acceleration signs is crucial in physics.
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A particle is moving with a velocity of 60.0 m/s in positive x direction at t= 0 and t = 15.0 s the velcity decreases uniformly to zero . What was the acceleration during this 15.0-s interval ? what is the significance of the sign of your answer ?





my answer is



Vo = 0

a = 60 - 0 / 15 = 4
 
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Hi r-soy! :smile:

You must get into the habit of writing the formula clearly on one line, then filling in the numbers on the next line …

if you'd done that, you wouldn't have got the wrong answer! :redface:

So try again, and start with the full formula. :smile:​
 

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