A train takes 10 mins to travel from A to B. The train accelerates at a rate of 0.5ms^ -2 for 30 secs.?

It then travels at a constant speed and is finally brought to rest in 15 secs with constant deceleration.
Find the constant speed
Find the rate of deceleration. .
Find the distance form A to B.

constant speed is 15 ms^-1
rate of deceleration is 1 ms^-2
distance form A to B is 15.075 km

t = 30s
a = 0.5
u=0
v=?
a=(v-u)/t v= 15m/s (constant speed)

t= 15
v=0
u=15
a=?
a=(v-u)/t a=-1ms^-2 (rate of deceleration)

(0.5x15x30) + (600 - 30 -15) x15 + (0.5x15x15)
= 8.66 km

obv 1st part of v/t graf has grad=0.5..so const v=0.5*30=15m/s..final sect grad=decel=15/15=-1m/s/s
use d=ave v*t summed ovr 3 sections
ie d=7.5x30+15x555+7.5x15=8662.5m

Sorry I don’t do anybody’s homeworks you cheater!