Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

 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

Grade:9

7 Answers

AKASH GOYAL AskiitiansExpert-IITD
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

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

 

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.

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.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..

 

 


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.

 

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

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.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar : Click here to download the toolbar.. 

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.

vikas askiitian expert
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
askIITians Faculty 646 Points
11 months 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

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free