Question

A particle is moving along x axis the position of particle is given by x= 20+0.1t ka square find 1st average acceleration of the particle between t=2s and t=3s .and 2nd show that acceleration of the particle is constant

Answer 1

Explanation:

here x=(20+0.1t)²

v=dx/dt=0.2(20+0.1t)

v2(t2=3s)=4.06m/s

v1(t1=2s)=4.04m/s

so average acceleration =V2-V1/t2-t1=0.02m/s²

also acceleration a=d²x/dt²=d/dt{0.2(20+0.1t)}

d/dt(4+0.02t)=0.02m/s²

so here acceleration is constant