Imaginary Can We Find the Least Time?

[chwala]

Homework Statement

See attached.

Relevant Equations

$s$=$ut$+$\frac {1}{2}$$at^2$

This is the problem;

I tried using;
$s$=$ut$+$\frac {1}{2}$$at^2$
and ended up with, $240$ =$12t -1.5t^2$ clearly we have an imaginary solution here and therefore this may not be correct?
My wild guess is that the least time (Without deceleration) should take at least, $t$=$\frac {240}{12}$$=20$seconds

 

[PeroK]

Homework Statement:: See attached.
Relevant Equations:: $s$=$ut$+$\frac {1}{2}$at^2##

This is the problem;

View attachment 294254I tried using;
$s$=$ut$+$\frac {1}{2}$$at^2$
and ended up with, $240$ =$12t -1.5t^2$ clearly we have an imaginary solution here and therefore this may not be correct?
My wild guess is that the least time (Without deceleration) should take at least, $t$=$\frac {240}{12}$$=20$seconds

Okay, if he doesn't brake, then he takes $20s$ to travel $240m$ at $12m/s$.

But, he needs to brake and come to a stop at $B$.

 

[chwala]

Okay, if he doesn't brake, then he takes $20s$ to travel $240m$ at $12m/s$.

But, he needs to brake and come to a stop at $B$.

True, I can't seem to find the solution. I have the textbook solution with me...I need input...I have taken a break just to refresh and think about this question...

 

[PeroK]

True, I can't seem to find the solution. I have the textbook solution with me...I need input...I have taken a break just to refresh and think about this question...

If you were the cyclist, what would you do?

 

[chwala]

If you were the cyclist, what would you do?

Of course decelerate...at which we have $v=0$ at point $B$. We can't use $v = u +at$, here as $t=4$, is misleading.

 

[PeroK]

Of course decelerate...at which we have $v=0$ at point $B$. We can't use $v = u +at$, here as $t=4$, is misleading.

When would you hit the brakes is the question?

 

[PeroK]

PS do you know about the concept of "stopping distance", as it relates to road vehicles?

 

[chwala]

When would you hit the brakes is the question?

I need to think about this...

 

[chwala]

aaaaaaaaaaaaaaaaaaaaaargh i got it! Let me post my working.

 

[chwala]

Now when the brakes are applied ( retardation)...leading to point $B$, this takes a total time of $t=4$ as earlier indicated in post $6$. Now distance traveled in the last $4$ seconds is,
$s$=$ut$+$\frac {1}{2}$$at^2$
$48-24= 24$ metres
Therefore the first portion of the journey takes $t$= $\frac {216}{12}=18$seconds.

Therefore the cyclist will travel atleast $18+4=22$ seconds. Phew!