Question

A commuter train travels between two stations by accelerating at the rate a for a time t1 and then slowing down with a negative acceleration of -5a for a time t2. The two stations are a distance d apart and the train never reaches its maximum speed. The train stops at each station. Find times t1 and t2.

Answer 1

I'm also not allowed to use any other equation except these:
\[v^{\rightarrow}=\frac{dr^{\rightarrow}}{dt}\]
\[a^{\rightarrow}=\frac{dv^{\rightarrow}}{dt}\]
\[v^{\rightarrow}=v _{0}^{\rightarrow}+at^{\rightarrow}\]
\[r^{\rightarrow}=r _{0}^{\rightarrow}+v _{0}^{\rightarrow}t+\frac{1}
{2}a^{\rightarrow}t^{2}\]

Answer 2

during time t1:
r - ro = vo*t1 + (1/2)*a*t1^2
but vo = 0
d1 = (1/2)*a*t1^2

and when t = t1:
v = a*t1

during time t2:
vo = v, from the previous step
r - ro = vo*t2 +(1/2)*(-5a)*t2^2
d2 = a*t1*t2 - (5/2)*a*t2^2

and when t = t2:
0 = a*t1 -5a*t2
t2 = (1/5)*t1

d1+d2 = d

d = (1/2)*a*t1^2 + a*t1*t2 - (5/2)*a*t2^2

plug in the relation between t2 and t1:

d = (1/2)*a*t1^2 + a*t1*t2 - (5/2)*a*t2^2

d = (1/2)*a*t1^2 + a*(1/5)*t1^2 - (5/2)*a*((1/5)*t1)^2

d = (1/2)*a*t1^2 + a*(1/5)*t1^2 - (1/10)*a*t1^2

(1/2 + 1/5 - 1/10)*a*t1^2 = d

(3/5)*a*t1^2 = d

t1^2 = (5/3)*d/a

t1 = sqrt((5/3)*d/a)

Now find t2:

t2 = (1/5)*t1

t2 = (1/5)*sqrt((5/3)*d/a)