Question

acceleration of a particle as given by a = 3t at the=0,v=0,x =0. The velocity at t=2 sec will be
(1) 6m/s , 4m
(2) 4 m/s , 6m
(3) 3 m/s, 2 m
(4) 2 m/s, 3m​

Answer 1

Answer:

Let’s begin with a particle with an acceleration a(t) is a known function of time. Since the time derivative of the velocity function is acceleration,

\[\frac{d}{dt}v(t)=a(t),\]

we can take the indefinite integral of both sides, finding

\[\int \frac{d}{dt}v(t)dt=\int a(t)dt+{C}_{1},\]

where C1 is a constant of integration. Since

\[\int \frac{d}{dt}v(t)dt=v(t)\]

, the velocity is given by

\[v(t)=\int a(t)dt+{C}_{1}.\]

Similarly, the time derivative of the position function is the velocity function,

\[\frac{d}{dt}x(t)=v(t).\]

Thus, we can use the same mathematical manipulations we just used and find

\[x(t)=\int v(t)dt+{C}_{2},\]

where C2 is a second constant of integration.