Question

Position vector ofa particle is given by T =
${3t}^{3} i + 4tj + {t}^{2} k$
.Find avg acceleration of particle from
t= 1 to t= 2sec.

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Answer 1

Answer:

Given x=(t−2)2=t2−4t+4 

Now at t=0

x=4 m

at t=2s

x=4−8+4=0 m

and at t=4 s

x=16−16+4=4 m

Thus total distance covered at t=4s=8 units.