Question

the acceleration of a particle a=3t^2+2t+2 in SI system. If the particle start with velocity, v=3m/sec at t=0, then find the velocity at the end of 2sec.​

Answer 1

Answer:

39m/s^2

Explanation:

u=3m/s, t=2sec, a=3t^2+2t+2

we know:

v=u+at

v=3+[(3×2×2)+(2×2)+(2)]2

v=3+36

v=39m/s

velocity after 2 sec= 39m/sec