A body at rest is given an initial uniform acceleration of 8m/s2 for 30s after which the acceleration is reduced to 5m/s2 for the next twenty seconds. The body maintains the speed attained for 60s, after which it is brought to rest in 20s.

Question

1. Draw the velocity time graph of the motion using the information given.
2. Using the graph, calculate the maximum speed attained during the motion
3. Average retardation as the body is being brought to rest
4. Total distance travelled during the first 50s
5. Average speed during the first 50s

Answer 1

i don't know the answer

Answer 2

answer

Answer 3

I did not know it
i seek for the answer

Answer 4

I dont know help me please

Answer 5

pls help i do not know

Answer 6

1.u=0
V=u+at=0+8*30=240

Answer 7

1)v1=u+a1t1=0+(8*3)=240m/s
2)v2=v1+a2t2=240+(-5)*20=140m/s
3)total area convered in 50s=Area of motion in 30s+Area of motion in 20s=area of triangle + area of parallelogram=1/2*30*240+1/2(240+140)*20=7400m
4)average speed=total distance covered in 50s/50s=7400/50=148m/s

Answer 8

I don’t know the answer