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Shivam Bhagat Grade: 9
        

 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


7jul10-img1.jpg


(A)    110 m/s


(B)    55 m/s


(C)    550 m/s


(D)    660 m/s

6 years ago

Answers : (6)

AKASH GOYAL AskiitiansExpert-IITD
419 Points
										

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


 


All the best.                                                           


AKASH GOYAL


AskiitiansExpert-IITD


 


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6 years ago
Kuldeep Rajput
31 Points
										

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

6 years ago
kedar joshi
35 Points
										

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

6 years ago
Aman Bansal
592 Points
										

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.


 


We are all IITians and here to help you in your IIT JEE preparation.


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AMAN BANSAL

6 years ago
ANKIT BATHLA
34 Points
										

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.

6 years ago
vikas askiitian expert
510 Points
										

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

6 years ago
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