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

Gold Member
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.

Mentor
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##.

chwala
Staff Emeritus