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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
```

6 years ago

Share

```										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

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AKASH GOYAL

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```
6 years ago
```										55.AREA UNDER acc.time graph change in velocity .
```
6 years ago
```										the max. velocity of the particle will be simply the area enclosed by the a-t graph ie 55m/s
```
6 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
```
6 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.
```
6 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
```
6 years ago

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