IIT JEE , AIEEE Forum - Physics , Mechanics

cutie_divs91

Starting from rest a particle moving in a straight line is accelarated with an acceleration ?

A=(16-t^2)^1/2 for 0<=t<=4s

A=2 m/s2 for t>4s

What is the velocity of the particle after 6 s?

a) 10.82

b) 12.50

c) 14.2

d) 16.56

elessar_iitkgp

a =

16-t²
₀

^v' dv =₀

⁴

16-t² dt
v' =₀

⁴

16-t² dt
Evaluate the integral and find v'
If velocity at t = 6s be V,
V = v' + 2(6-4)

Put the value of v' and get the answer