* A train accelerates uniformly from rest at point 0 to 30 m/s at point A.
The distance travelled during this stage is 900 m.
At point A, the train continues with a constant velocity until reaching
point B. The distance travelled during this stage is 1500 m.
Determine the average velocity of the train (V OB=

21.82 m/s

Explanation:

Given that :

Point OA:

Initial Velocity (u) = 0

Final velocity (v) = 30 m/s

Distance OA = 900m

Time taken to cover distance OA:

Using the motion equation:

Acceleration(a) :

(v - u)² / 2(s)

(30 - 0)² / 2*900

900 / 1800

= 0.5m/s²

From :

v = u + at

toa = (v - u) / a

(30 - 0) / 0.5

30/ 0.5 = 60 seconds

Time taken to cover OA = 60 seconds

Distance AB:

Distance traveled = 1500m

Velocity of travel = constant = 30m/s

Time taken = distance / velocity

Time taken = 1500 / 30

Time taken = 50 seconds

Hence, average velocity (Vob)

Total distance covered / total time taken

(900 + 1500)m / (60 + 50)s

2400 m / 110 s

= 21.818 m/s

= 21.82 m/s

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6