The maximum rate at which a bus can accelerate or decelerate is 2m/s^2 it has a maximum speed of 60km/h. find the shortest time the bus can take to travel between two bus stops 1 km apart on a straight stretch of road.

Question did not specify that bus was stationary at any point.

60 km/h = 60000/3600  m/s = 16.67 m/s
Shortest time to travel = distance / speed = 1000/(50/3)=60 sec. = 1 min. 

If we assume bus was stationary at each stop (if passengers have to get on/off), then acceleration is involved.

Note: at 2 m/s, the bus accelerates from stationary to 60 km/h in 8 seconds!  

Using kinematics equation
2aS=v^2-u^2
we have
a=acceleration = 2m/s
S=distance
u=0 initial speed
v=final speed=50/3 m/s=60 km/h
Distance travelled = (v^2-u^2)/(2a)=(50/3)^2/(2*2)=625/9 m=69.33 m.
Remaining distance at 60 km/h
=1000-69.44-69.44=861.11 m
time to travel = 861.11/(50/3)=51.67 s.
Time to accelerate/decelerate= (50/ 3)/2 m/s /  m/s^2 = 8.333 s.
Total time = 51.67+2*8.33=68.33 s.

You have 2 sides that are 2 cms and you have 6 sides that are 1/2 cm and you have 4 sides that are 1 cm. the smallest sides are 1/2 the medium sized sides are 1 and the smallest are 1/2. so to find the perimeter you have to add them all up together so your answer will be 11 cm's

3units/ 5 feet = 60 units/ 100 feet3 units/ 5 feet = 18 units/ 30 feetthe dimensions of the scale drawing are 60 units by 18 units.hope this !