 # A particle starts form rest. Its acceleration (a) versus time (t) is as shown in the figure. The maximum speed of the particle will be (A)    110 m/s(B)    55 m/s(C)    550 m/s(D)    660 m/s AKASH GOYAL AskiitiansExpert-IITD
419 Points
11 years ago

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

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

All the best.

AKASH GOYAL

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11 years ago

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

11 years ago

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.

11 years ago

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

2 years ago
Dear Student,
Please find below the solution to your problem.
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.

Thanks and Regards