SOLUTION: a cart is moving at a initial velocity of 2.0 m/s and accelerates at 3.3 m/s^2 for 33 m . how fast is it going at the end of the motion

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Question 1123033: a cart is moving at a initial velocity of 2.0 m/s and accelerates at 3.3 m/s^2 for 33 m . how fast is it going at the end of the motion
Found 3 solutions by ikleyn, josmiceli, Alan3354:
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
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Let me apply the Energy conservation law.


The change of kinetic energy is  %28mV%5E2%29%2F2 - %28mV%5B0%5D%5E2%29%2F2.


The work of the force which provided the given acceleration "a", is  (ma)*d,

where d is the distance (of 33 m in your case).


So, the conservation of energy gives an equation


    %28mV%5E2%29%2F2 - %28mV%5B0%5D%5E2%29%2F2 = ma*d.


Cancel the mass "m" in both sides.  You will get


    %28V%5E2%29%2F2 - %28V%5B0%5D%5E2%29%2F2 = a*d.


Hence,  V%5E2%2F2 = ad + %28V%5B0%5D%5E2%29%2F2, 

and finally


    V = sqrt%28V%5B0%5D%5E2+%2B+2ad%29.


Now substitute the given values into the formula and calculate the answer.


It is, probably, the shortest way to solve the problem and to explain the solution.


Did I pass your exam ?

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What is your textbook in Physics ?



Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+d+=+v%5B0%5D%2At+%2B+%281%2F2%29%2Aa%2At%5E2+
+v%5B0%5D+=+2+ m/sec
+a+=+3.3+ m/sec2
+d+=+33+m
+33+=+2t+%2B+%281%2F2%29%2A3.3%2At%5E2+
+33+=+2t+%2B+1.65%2At%5E2+
+165t%5E2+%2B+200t+-+3300+=+0+
+33t%5E2+%2B+40t+-+660+=+0+
Solve for t using quadratic formula
Let +v%5Bf%5D+ = final velocity
+%28+v%5Bf%5D+-+2+%29+%2F+t+=+3.3+

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
a cart is moving at a initial velocity of 2.0 m/s and accelerates at 3.3 m/s^2 for 33 m . how fast is it going at the end of the motion
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Is it 33 meters, or 33 minutes?