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Mathematics 6 Online
OpenStudy (anonymous):

Find the distance travelled by the car if it starts moving from rest and has been travelling for 7 seconds, with the maximum acceleration of 6ms^-2.

OpenStudy (anonymous):

Is the acceleration constant?

OpenStudy (anonymous):

It's not mentioned.

OpenStudy (anonymous):

\[x=1/2(6)7^2\]

OpenStudy (anonymous):

Assuming it is: x=ut+0.5*a*t^2 u=0 x=0.5*6*7^2 x=147m

OpenStudy (anonymous):

No, I don't think that's the right way. It's just the max. acceleration.. And this question is from a math quiz, carrying 25 marks.

OpenStudy (anonymous):

\[y=0+1/2(at^2)\] \[a=\frac{ 2y }{ t^2 }\]\[.........................a^'=\frac{ 2y^'t^2-4ty }{ t^4 }\] \[y=\frac{ v }{ 2t }\] \[............................y^'=2v \ln t\] max a \[0=\frac{ 4vt \ln t -2v}{ t^4 }\] \[t\ln t=\frac{ 1 }{ 2 }\] \[t^t=0,5\]

OpenStudy (anonymous):

CRAZY

OpenStudy (experimentx):

\[ s = \int_{0}^{7} v(t) dt \\ v(t) = \int_0^7 a(t) dt \\ s = \int_0^7 \int_0^7 a(t)\; dt \; dt \]

OpenStudy (anonymous):

and the equation of a(t) is....?

OpenStudy (experimentx):

all we know is \( |a(t)| \leq 6 \)

OpenStudy (experimentx):

distance can be expressed in terms of inequality ... the distance depends on a(t) so we do not know exactly.

OpenStudy (experimentx):

one example of such acceleration would be \[ a(t) = {-t(t-7)24 \over 49 } \]

OpenStudy (anonymous):

But we don't know what a(t) is, how can we integrate that?

OpenStudy (experimentx):

you can set a = 6 and express it in terms of inequality. there is no certainty ... is it stat question?

OpenStudy (anonymous):

Calculus.

OpenStudy (experimentx):

do you have answers?

OpenStudy (anonymous):

no.

OpenStudy (experimentx):

probably ... answer is in range [from this distance to this distance]

OpenStudy (anonymous):

Would it be something related to parametric equations?

OpenStudy (experimentx):

ah sorry this would be this \[ s = \int_{0}^{7} v(t) dt \\ v(t) = \int_0^t a(t) dt \\ s = \int_0^7 \int_0^t a(t)\; dt \; dt \]

OpenStudy (mayankdevnani):

@RolyPoly do you know the second eq. of motion

OpenStudy (mayankdevnani):

do you know?

OpenStudy (mayankdevnani):

@RolyPoly

OpenStudy (anonymous):

Under constant acceleration, I know. But sorry that we don't even know if the motion is of constant acceleration!

OpenStudy (mayankdevnani):

we can suppose....

OpenStudy (anonymous):

We can't! It's a MATH question after all!~

OpenStudy (mayankdevnani):

ok sry

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