Dear Shivam

From the graph you can see that acceleration is decreasing but it is always positive and hence its direction is same throughout from t=0 to t=11 and velocity will keep on increasing and will be in the direction of acceleration. hence magnitude of velocity can be taken as speed. and maximum speed will be at t=11 sec

a=dv/dt

dv= adt

integrating from t=0 to t=11

change in velocity=v₂-v₁ area under a-t graph=1/2 x 11 x 10 =55m/s

since v₁ is zero then v₂=55 m/s

(B) answer

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AKASH GOYAL

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55.AREA UNDER acc.time graph change in velocity .

the max. velocity of the particle will be simply the area enclosed by the a-t graph ie 55m/s

DEAR SHIVAM BHAGAT,

we can evaluate the equation of line as follow,

 10t + 11a = 110 

then we can write a= dv / dt,

thus we can evalute the equation for velocity as

11v = 110 t - 5t^2

and its value is maximum at t= 11 s ( this can be evaluated by differentiating and equating to 0 )

thus we calculated maX V  AS  55 m/s.

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AMAN BANSAL

here a=10/-11(t-11) =dv/dt=-10/11t+10 formaxima dv/dt=0 so -10/11t=-10so/ t=11 now on integration we get v=-10t²_{/22+10t     now put t=11 we get 55 as answer.}

eq of straight line in the graph is

(Y-O)=-10/11 (X-11)

 Y = -10(x-11)/11

now , y represents a & x represents t

a = -10(t-11)/11            ...............1

V = adt = -10(t²/2-11t)/11              .............2

for maximum speed  , put a=0 , so

t = 11sec ,

at t = 11 , velocity will be maximum

V = -10(121/2 - 11*11)/11

V = 55m/s