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Mathematics: Post your doubts here!

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Part 2 .Lil Confuse . Please do the solution on paper and post the snapshot here.


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and Part 3

Suchal Riaz
 
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the thing that's confusing me here is that after differentiating we get t=100 ...
isn't t=100 the time at which the speed is max?
in the ms they have considered it to be the the time taken for the distance AB travelled.... it's confusing me to bits!!:/

and umm well solve the rest of the questoin too if u can :)
 
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View attachment 39978

the thing that's confusing me here is that after differentiating we get t=100 ...
isn't t=100 the time at which the speed is max?
in the ms they have considered it to be the the time taken for the distance AB travelled.... it's confusing me to bits!!:/

and umm well solve the rest of the questoin too if u can :)
Did you open the bracket when differetiating ? you should do that !
 
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Did you open the bracket when differetiating ? you should do that !
yeah i did get the right answer... it was t=100

what i'm asking is that is t=100 the time when the particle reaches B??
shouldn't 100 be the time when the speed is max??
 
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rearrange the equation such that it is in the form:
lny = lnA - kx^2
-->
lny = -kx^2 + lnA

from this u can see that ur gradient will be -k and ur y-intercept is lnA

find the gradient using the coordinates of the two given points and equate it to -k .... then put in the values of lny , k, and x^2 from one of the points in the equation and find c i.e. lnA

got it?
 
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yeah i did get the right answer... it was t=100

what i'm asking is that is t=100 the time when the particle reaches B??
shouldn't 100 be the time when the speed is max??
when we differtiate equation we get Velocity equation
so to find time when V=0 , we equate zero with the equation of velocity , so we get t=0 & t=100
question says clearly v=0 , when start at rest and when reach B , read the question , so 0 is time at A , and 100 is time at B
 
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rearrange the equation such that it is in the form:
lny = lnA - kx^2
-->
lny = -kx^2 + lnA

from this u can see that ur gradient will be -k and ur y-intercept is lnA

find the gradient using the coordinates of the two given points and equate it to -k .... then put in the values of lny , k, and x^2 from one of the points in the equation and find c i.e. lnA

got it?
X^2 + 5X = 8 in the form of (ax^2 +bx) +C
Pls help. :)
 
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