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# 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

419 Points
10 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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Kuldeep Rajput
31 Points
10 years ago

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

kedar joshi
35 Points
10 years ago

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

Aman Bansal
592 Points
10 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.

Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

AMAN BANSAL

ANKIT BATHLA
34 Points
10 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.

509 Points
10 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

Rishi Sharma
11 months ago
Dear Student,
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