1.

The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity v_0. The distance travelled by the particle in time t will be

Answer»

`nu_(0)t+(1)/(3) BT^(2)`
`nu_(0)t +(1)/(3) bt^(3)`
`nu_(0)t+(1)/(6)bt^(3)`
`nu_(0)t+(1)/(2)bt^(2)`

Solution :Acceleration `a= (dv)/(dt) = bt` , so , dv= BTDT
Integrating equation (i) within the conditions of motion we have ,
`int_(nu_(0))^(nu)dnu=int_(0)^(t)"btdt or "(nu-nu_(0))=(bt^(2))/(2)`
or `nu=nu_(0)+(bt^(2))/(2)=(ds)/(dt) " or ds " = nu_(0)dt +(bt^(2))/(2) `dt
Integrating equation (II) we get s =` nu_(0)t +(1)/(6) bt^(3)`


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