rst
Youngling
Average velocity= 72 km/h= 20m/s
Total time= 25s
Total distance= 20*25=500m
Let t1,t2,t3 be time for acceleration, uniform motion and retardation
Time for acceleration and retardation will be same
So t1 +t2+t1= 25
2 t1 + t2 = 25
Velocity after time t1
v1 = 0+5t1
Now we can draw v- t gragh of above motion which will be like trapezium
Distance= area of trapezium
500= 1/2 (t2+25)v1
1000= (25-2t1+25) 5t1
Solving t1=5
So t2=15s
Total time= 25s
Total distance= 20*25=500m
Let t1,t2,t3 be time for acceleration, uniform motion and retardation
Time for acceleration and retardation will be same
So t1 +t2+t1= 25
2 t1 + t2 = 25
Velocity after time t1
v1 = 0+5t1
Now we can draw v- t gragh of above motion which will be like trapezium
Distance= area of trapezium
500= 1/2 (t2+25)v1
1000= (25-2t1+25) 5t1
Solving t1=5
So t2=15s
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