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`        a particle starts from rest. its acceleration verses time graph is given.the maximum speed of the particle is. the graph is a right angled triangle with y axis as the height and x axis as the base.it is an acceleration time graph.the highest acceleration of the particle is 10m/s^2..it touches the x axis at time = 11 	110m/s	55m/s	550m/s	660m/s`
5 years ago

```							find the relation between a and t the above is a stright line so  appling intersept form of line we get (a/10)+(t/11)=1now a=10(1-(t/11))now put a=dv/dt and find v the velocity is maximum wn a=0 so at t=11so we get vmax=55PLZ APPROVE
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5 years ago
```							incase of any doubts  ask in ans box
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5 years ago
```							can u make it more clearhow can velocity be maximum when t=11 as the body is retarding
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5 years ago
```							given any function f(x) then the maximum or minimum values of function lies at “x” where  df(x)/dx=0here f(x) is v (velocity) and   df(x)/dx is dv/dt which is a (XLR8n)so v is maximum when a=0 so a is zero at t=11  PLZ APPROVE if useful in case of any doubts ask in ans box i will b avaible after 7:30
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5 years ago
```							the body is not retading it will retard after t=11
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5 years ago
```							As the Graph has Slope down straight line.Therefore we can apply slope intercept form.(a/10) + (t/11) =1Solve and we will get a =10-10t/11.................. Eqn 1As, a=dv/dt............(use it un eqn 1)We get ,dv/dt=10-10t/11dv=10dt - 10tdt/11Integrate and we get...V=10t-10(t^2)/11*2Now ,V will ve max at t=11 sec.Therefoe V=110- 5*121/11V=110-55v=55 m/s. Is the answer
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3 years ago
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