Math Is Fun Forum

a=2/3 is the answer!
Note: t1 t2 s1 s2 are the times and distance covered in acceleration and deceleration process.

v=0+at1  and   0=v-3at2   implies t1=3t2 also s/(t1+t2)=sqrt(s/2) which gives t1+t2=sqrt(2s)  putting t1=3t2 we get 4t2=sqrt(2s)  and t2^2=s/8

now, v^2=2as1 and 0=v^2-6as2 implies s1=3s2 also s1+s2=s putting s1=3s2 gives 4s2=s i.e s2=s/4

we had v=3at2(see 3rd line) so v^2 =9(a^2)(t2^2)=6as2(see 4th line) now putting t2^2=s/8(see 3rd line) and s2=s/4(see 4th line)hm in this eqn we get a=2/3