What is the least time to get from point ##A## to point ##B##

[chwala]

Homework Statement

A cyclist travels from point $A$ to $B$, a distance of $240$ metres. He passes $A$ at $12$ m/s, maintains this speed for as long as he can, and then breaks so that he comes to a stop at $B$. If the maximum deceleration he can achieve when braking is 3 m/s^2, what is the least time in which he can get from $A$ to $B$?

Relevant Equations

Velocity and acceleration

My approach;

$v=u+at$

$0=12-3t$

$t=4$

i.e at point when deceleration starts up to the point cyclist stopped (point $B$).

Therefore, distance travelled in the $4$ seconds is given by,

$s=(12×4)+(0.5×-3×16)=48-24=24$m

$⇒240-24=216$m

$t=\dfrac{216}{12}=18$seconds.

Therefore least time taken is $4+18=22$ seconds.

Text book has only given the answer $22$

your insight or alternative approach is welcome.

 

[DrClaude]

Your approach is correct and I would have done it the same way.

Two things I don't appreciate: the lack of units in the intermediate equations and the fact that you have not stated the equation for $s$ you are using. Also $0.5 \times -3 \times 16$ is not proper mathematics, that should be $0.5 \times (-3) \times 16$.

 

[vela]

Try analyzing the cyclist's motion using a velocity vs. time graph.