a particle covers 10m in first 5sec and 10m in next 3sec assuming constant acceleration find initial speed acceleration and distance covered in next 2sec
shuttlersaiyam:
Please elaborate
Please explain. I got the thing taking the speed 20/8 at the second point.
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given time = 5 sec
distance = 10 m
∴speed = 2m/s
again,
time = 8 sec
distance = 20 m
∴ speed = 20/8 m/s
∴ using v = u + at
20/7 = 2 + 8a
a = 6/ 64 m/s²
∴ in time 10 sec it will cover distance
s = ut +1/2at²
s = 28.125 m
∴ distance covered in next 2 sec
28.125 m - 20 m = 8 m (approximately)
distance = 10 m
∴speed = 2m/s
again,
time = 8 sec
distance = 20 m
∴ speed = 20/8 m/s
∴ using v = u + at
20/7 = 2 + 8a
a = 6/ 64 m/s²
∴ in time 10 sec it will cover distance
s = ut +1/2at²
s = 28.125 m
∴ distance covered in next 2 sec
28.125 m - 20 m = 8 m (approximately)
When you apply v=u+at, how the v comes to be 20/7 when in the step just before it, it is 20/8
ok
Huj? Didnt get you?
it is
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