There are 3 parts to the motion

part 1 takes 10 s

part 2 takes 20 s

and part 3 takes 5 s.

Total time taken is 35 s

To calculate the average velocity, we must work out the distance covered.

Part 1: s = ut + 0.5*a*t^2

s = 0 + 0.5*2*10^2

s = 100 m

Part 2: s = v*t = 20*20 = 400 m

Part 3: First calculate the deceleration, then the distance

v = u + at

0 = 20 + a*5

a = -4 m/s^2

s = ut + 0.5*a*t^2

s = 20*5 - 0.5*4*5^2 = 50m

Total distance travelled is 550 m

average velocity = distance/time = 550/35

v = 15.7 m/s

There are 3 parts to the motion

part 1
t 1=( v-u) /a
t1 = 10 s

part 2

Given that t2 = 20s.
part 3
Given that t3 = 5 s.
Total time taken is 35 s

To calculate the average velocity; we must work out the total distance covered.

Part 1:
Average velocity is [20 +0] / 2 = 10
s1 = 10 x 10 = 100 m

Part 2:
s2 = v*t = 20*20 = 400 m
Part 3:
Average velocity is [20 +0 ] / 2 = 10

S3 = 10 x5 = 50 m

Total distance traveled is 550 m

average velocity = distance/time = 550/35

v = 15.7 m/s

first v=u+at where

where v - final velocity a-acceleration t- time u- initial vel (zero in this case)
since the rate of acceleration (a) is same till it reaches 20m/s

time taken to reach 20m/s with an acceleration of m/s2 is 10 sec

later it has travelled for 20 sec at constant speed and another 5 sec to come to rest

over all it took 35sec travel.

how far has it travelled??
s=ut+0.5*a(t^2)
as u = 0
s=100m
i.e b4 reaching 20m/s it has travelled 100m
and at 20m/s it has travelled for 20 sec so tat makes it 400m
aftet that it travels for 5 sec

as v=u+at after that 5 sec v=0
hence 0=20+a5
thnce a=-4m/sec

s(distance) = ut+0.5a (t^2)
as u = 20m/s
a= -4m/s2
t=5
s=50m

hence it has travelled a total distance of 100+400+50 = 550m

to calculate the average velocity it has travelled 550m in a total time of 35sec
hence it is 550/35 = 15.71 m/s