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Archita Grade: 11
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^ touches the x axis at time = 11
  1. 110m/s
  2. 55m/s
  3. 550m/s
  4. 660m/s
2 years ago

Answers : (6)

Nicho priyatham
626 Points
find the relation between a and t 
the above is a stright line so  appling intersept form of line we get 
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
2 years ago
Nicho priyatham
626 Points
incase of any doubts  ask in ans box
2 years ago
132 Points
can u make it more clear
how can velocity be maximum when t=11 as the body is retarding
2 years ago
Nicho priyatham
626 Points
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 
2 years ago
Nicho priyatham
626 Points
the body is not retading it will retard after t=11
2 years ago
Himanshu Chaubey
13 Points
										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
6 months ago
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