Question

A particle is moving along x axis . the position of particle at any instant is given by x=10+0.2t^2, where x is measured in meters and time in second . The average acceleration of the particle between t=2sand t=3s

Answer 1

Answer:

we can take the indefinite integral of both sides, finding

∫ddtv(t)dt=∫a(t)dt+C1,∫ddtv(t)dt=∫a(t)dt+C1,

where C1 is a constant of integration. Since ∫ddtv(t)dt=v(t)∫ddtv(t)dt=v(t), the velocity is given by

v(t)=∫a(t)dt+C1.v(t)=∫a(t)dt+C1.

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

ddtx(t)=v(t).ddtx(t)=v(t).

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

x(t)=∫v(t)dt+C2,x(t)=∫v(t)dt+C2,

where C2 is a second constant of integration.

We can derive the kinematic equations for a constant acceleration using these integrals. With a(t) = a a constant, and doing the integration in (Figure), we find

v(t)=∫adt+C1=at+C1.v(t)=∫adt+C1=at+C1.

If the initial velocity is v(0) = v0, then

v0=