A car accelerates from 10 m/s to 45 m/s in 12 seconds. Calculate its acceleration.

Aparticle with mass 1.81×10−3 kg and a charge of 1.22×10−8 c has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. what are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field b⃗ =(1.63t)i^+(0.980t)j^?

Did the proton move into a region of higher potential or lower potential? did the proton move into a region of higher potential or lower potential? because the proton is a negative charge and it accelerates as it travels, it must be moving from a region of higher potential to a region of lower potential.because the proton is a negative charge and it accelerates as it travels, it must be moving from a region of lower potential to a region of higher potential.because the proton is a positive charge and it slows down as it travels, it must be moving from a region of higher potential to a region of lower potential.because the proton is a positive charge and it slows down as it travels, it must be moving from a region of lower potential to a region of higher potential.request answerpart bwhat was the potential difference that stopped the proton? express your answer with the appropriate units.î”v î” v = nothingnothingrequest answerpart cwhat was the initial kinetic energy of the proton, in electron volts? express your answer in electron volts.ki k i = nothing ev request answerprovide feedback

Given the position vector of the particle r(t)=(t+1)i+(t^2−1)j+2t k, find the particle's velocity and acceleration vectors at t=1