DYNAMICS
posted by HELP .
A particle travels along a straight line with a velocity v=(123t^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

v=123t^2
acceleration= 6t at t=4, acceleration is 24m/s^2
displacement=INTEGRAL velocity dt
= INT(123t^2)dt= 12tt^3 from t=0 to 10
= 1201000=860