Rectilinear Motion- Car- Non constant accel

[ku1005]

Car starts at rest, initial acceleration = 4.6m/s^2.Accel decreases linearly with time to 0 in 12 seconds.After this the car travels at constant speed. Det the time required for the car to travle 546m from the starting position.

I did

a = 4.6-(4.6/12)*t (1) (ie straight line/linear relationship for 0<t<12)

thus given a = dv/dt

then

a dt = dv

hence integrating (1) above to find v

gives

v = 4.6t-4.6t^2/24


now since

v dt = ds

do the same again

except this time we have the limits of integration for ds as 0 an 546, and the limits for t = o and t (ie unknown time)

SOlving the cubic equation gives 3 answers

70.27,11.93,-10.20

I selected 11.93, BUT im wrong and was wodnering why??


Any tips/advice would be great!!!!!!

thanks for you time

 

[ku1005]

not to worry guys/girls

as always it wasnt the approach which failed, yet the actual simple arithmatic!!!!!...didnt integrate t properly hahah

thanks

 

[stapel]

it wasnt the approach which failed, yet the actual simple arithmatic!

It's so often the "simple" stuff that'll "get" you! :wink:

Glad to hear you figured it out. Good work!

Eliz.