- #1

chwala

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- 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.

your insight or alternative approach is welcome.

##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.