V=u+at = 0+6(20)=120m/s
let the total duistance covered be represented by A
A= 1/2(a+b)h
A=1/2(20+60)120
A=1/2*80*120
A= 4800m=4.8km

Why did you multiply again by 20( the height)?

t1 = 20 sec  
t2 = 20 secs  
t3 = 20 secs      
Formular for total distance is 1/2(a + b)*h 
where a = time of constant acceleration (t1) 
while b = total time of journey (t1 + t2 + t3)
h = highest velocity which is v = at
a = 20 
b = 60 
h = 6*20 = 120m/s 
Therefore, 1/2(20 + 60)*120 = 4.8km

<p>A train starting from rest accelerates at the rate of 6 m/s\(^2\) for 20 seconds to attain a constant speed and it further travelled for another 20 seconds and decelerates at rate of 3 m/s\(^2\) for 20 seconds. Calculate the total distance (in kilometre) traveled by the train.</p>

Earlier discussion