Answer to Question #79641 in Physics for Somya kumari

Question

If the position is given by
X=2t+3t^2+4t^3
1)find the velocity at time t=2second
2)The acceleration at time t=3 second

Answer 1

The velocity
v(t)=x^' (t)=2+6t+12t^2
v(2)=2+6×2+12×2^2=62 m/s
The acceleration
a(t)=v^' (t)=6+24t
a(3)=6+24×3=78 m/s^2