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`        The relation between time t and distance x for a particle moving in a straight line is t=ax2 +bx where a and b are constants. The reatardation is ?  `
2 years ago

anemous
117 Points
```							 we know that al=vdv/dx             v=velocity,al=acceleration,x=displacementfirst differentiate given eq.w.r.t. t1=2axdx/dt+dx/dt1=2axv+bv(1)               v=1/2ax+bdifferentiate(1)  w.r.t. t0=2av2+2axal+bal .(2)substitute v=1/2ax+b in (2)2a/(2ax+b)2+(2ax+b)al=0(2ax+b)al= -2a/(2ax+b)2          a|=-2a/(2ax+b)3
```
2 years ago
noogler
489 Points
```							@ ajay kotwal y r u copying my post?????????
```
2 years ago
noogler
489 Points
```							i posted this complete ans to him
```
2 years ago
noogler
489 Points
```							 retardation is nothing but negative acceleration.we know that v=dx/dt,al=dv/dt             v=velocity,al=acceleration,x=displacementfirst differentiate given eq.w.r.t. t1=2axdx/dt+dx/dt1=2axv+bv …..................(1)               v=1/2ax+bdifferentiate(1)  w.r.t. t0=2av2+2axal+bal .....................(2)substitute v=1/2ax+b in (2)2a/(2ax+b)2+(2ax+b)al=0(2ax+b)al= -2a/(2ax+b)2          al=-2a/(2ax+b)3                                        -ve sign indicates decrease in acceleration
```
2 years ago
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