a particle starts from a point with velocity of +6.0 m/S and moves with an acceleration of- 2.0m/s2 .show that after 6s the particle will be at starting point
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v=0
u = 6 m/s
a= -20 m/s ²
v=u +at
0 = 6 -20 t
- 6 = -20 t
t = 6 /20
t = 0.3 sec
it will take 0.3 sec to stop if it is retarding . now suppose it again starts acceleration backward with same velocity .
Then it reaches at the same spot before 6s
a= -20 m/s ²
v=u +at
0 = 6 -20 t
- 6 = -20 t
t = 6 /20
t = 0.3 sec
it will take 0.3 sec to stop if it is retarding . now suppose it again starts acceleration backward with same velocity .
Then it reaches at the same spot before 6s
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