v=12-3t^2
acceleration= -6t at t=4, acceleration is -24m/s^2
displacement=INTEGRAL velocity dt
= INT(12-3t^2)dt= 12t-t^3 from t=0 to 10
= 120-1000=-860
A particle travels along a straight line with a velocity v=(12-3t^2) m/s, when t is in seconds. When t=1s, the particle is located 10m to the left of the origin. Determine the accelaration when t=4s, the displacement from t=0 to t=10s, and the distance the particle travels during this time period..
Thank you for the answer of my 1st question..hope you could help me w/ this one.tnx
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