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A particle is moving under constant acce...

A particle is moving under constant acceleration `a=alphat+betat^(2)`, where alpha beta constants. If the position and velocity of the partical at start, i.e. `t=0` are `x_(0)` and `v_(0)`, find the displacement and velocity as a function of time t.

A

`x(t)=x_(0)+v_(0)t+1/6alphat^3+1/12betat^(4)`

B

`x(t)=x_(0)+v_(0)t+1/6alphat^(2)+1/(24)betat^(3)`

C

`x(t)=x_(0)+v_(0)t+1/(12)alphat^(2)+1/6betat^(3)`

D

`x(t)=x_(0)+v_(0)t+1/6alphat^(2) +1/12betat^(3)`

Text Solution

Verified by Experts

The correct Answer is:
A
`a=(dv)/(dt)=alphat+betat^2 " or " dv = (alphat +betat^2)dt`
Integrating it , we get
Integrating it again ,we get `int_(v_0)^(v)dv = int_(0)^(t)(alphat + betat^2)dt`
`v-v_0 =1/2 alphat^2+1/3betat^3,v=v_0+1/2alphat^2+1/3betat^3`
`(dx)/(dt)=v_0+1/2at^2+1/3betat^3,dx=(v_0+1/2alphat^2+1/3betat^3)dt`
Integrating it again , we get
`int_(x_0)^xdx=int_0^t(v_0+1/2alphat^(2)+1/3betat^3)dt`
`x - x_(0) =v_(0) t + 1/6 alphat^(3) + 1/12 beta t^(4)`
or `x = x_(0) + v_(0)t + 1/6 alpha t^(3) + 1/12 beta t^(4)`
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