Relate position, velocity, speed, and acceleration using derivatives

s given by my friend pls solve (Im pre foundation so I dont understand shit here)\newlineON INTEGRATION\newline2020. The acceleration aa in ms2\text{ms}^{-2}, of a particle is given by a=3t2+2t+2a=3t^{2}+2t+2, where tt is the time. If the particle starts out with a velocity v=2ms1v=2\text{ms}^{-1} at t=0t=0, then the velocity at the end of 2s2\text{s} is\newline(a) 12ms112\text{ms}^{-1}\newline(b) 14ms114\text{ms}^{-1}\newline(c) 16ms116\text{ms}^{-1}\newline(d) ms2\text{ms}^{-2}00\newline2121. A point Initially at rest moves along X-axis. Its acceleration varies with time as ms2\text{ms}^{-2}11. If it\newline2222. If the velocity of a particle is ms2\text{ms}^{-2}22, Where ms2\text{ms}^{-2}33 and ms2\text{ms}^{-2}44 are constants, then the distance travelled by It between is and 2s2\text{s} is\newline(a) ms2\text{ms}^{-2}66\newline(b) ms2\text{ms}^{-2}77\newline(c) ms2\text{ms}^{-2}88 (d) ms2\text{ms}^{-2}99\newline2323. The acceleration of a particle Increasing linearly with time a=3t2+2t+2a=3t^{2}+2t+200. The particle starts from origin with an Initial velocity a=3t2+2t+2a=3t^{2}+2t+211. The distance travelled by the particle In time tt will be\newline(a) a=3t2+2t+2a=3t^{2}+2t+233\newline(b) a=3t2+2t+2a=3t^{2}+2t+244\newline(c) a=3t2+2t+2a=3t^{2}+2t+255\newline(d) a=3t2+2t+2a=3t^{2}+2t+266\newline2424. The velocity of a particle is a=3t2+2t+2a=3t^{2}+2t+277. If its position a=3t2+2t+2a=3t^{2}+2t+288, at t=0t=0, then its displacement after unit time tt00 is\newline(a)tt11\newline(b) tt22\newline(c) tt33\newline(d) tt44\newline2525. A particle moving along x-axis has acceleration aa, at time tt, given by tt77, here tt88 and tt99 are constants. The particle at t=0t=0 has zero velocity. In the time interval between t=0t=0 and the instant when v=2ms1v=2\text{ms}^{-1}22, the particle velocity v=2ms1v=2\text{ms}^{-1}33 is\newline(a)v=2ms1v=2\text{ms}^{-1}44\newline(b) v=2ms1v=2\text{ms}^{-1}55\newline(c) v=2ms1v=2\text{ms}^{-1}66\newline(d) v=2ms1v=2\text{ms}^{-1}77\newline2626. A particle located at v=2ms1v=2\text{ms}^{-1}88 at time t=0t=0, starts moving along the positive x-direction with a velocity t=0t=000 that varies as t=0t=011. The displacement of the particle varies with time as\newline(a) t=0t=022\newline(b) t=0t=033\newline(c) t=0t=044 (d) t=0t=055\newline2727. The velocity of particle moving in the positive direction of X-axis varies as t=0t=066, where t=0t=077 is a positive constant. Assuming that at moment t=0t=0, the particle was located at the point v=2ms1v=2\text{ms}^{-1}88. Find (i) the time dependence of the position (ii) the time dependence of the velocity of the particle. (ii) the mean velocity of the particle averaged over the time that the particle takes to cover first 's' meters of the path [Ans: (i) 2s2\text{s}00 (ii) 2s2\text{s}11 (iii) 2s2\text{s}22]
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