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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

1. 110m/s
2. 55m/s
3. 550m/s
4. 660m/s
4 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)=1
now a=10(1-(t/11))
now put a=dv/dt and find v
the velocity is maximum wn a=0 so at t=11
so we get vmax=55
PLZ APPROVE

4 years ago

incase of any doubts  ask in ans box
4 years ago

can u make it more clear
how can velocity be maximum when t=11 as the body is retarding
4 years ago

given any function f(x)
then the maximum or minimum values of function lies at “x” where  df(x)/dx=0
here 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
4 years ago

the body is not retading it will retard after t=11
4 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

2 years ago
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