A body starts from rest with an acceleration of 4cm/sec^2.What will be its velocity after travelling a distance of 50c.m. How much distance shall be in 10 seconds after 50c.m point.
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Here starts your solution my friend—
u = 0
a = 4cm/sec^2
s = 50 cm
Using v^2 = u^2 + 2as
v^2 = 0 + 2(4)(50)
v^2 = 400
v = under root 400
v = 20 cm/sec
Therefore, the velocity after traveling a distance of 50 cm will be 20 cm/sec.
Now,
t = 10sec
u = 20 cm/sec
a = 4 cm/sec^2
Using s = ut + 1/2at^2
s = 20(10) + 1/2(4)(10^2)
s = 200 + 1/2(4)(100)
s = 200 + 2(100)
s = 200 + 200
s = 400 cm
Total distance = 400 + 50
= 450 cm
Therefore, the distance travelled in 10 seconds after the 50 cm point will be 400 cm.
Hope it helps........
Please mark me as the brainliest!!!!!
u = 0
a = 4cm/sec^2
s = 50 cm
Using v^2 = u^2 + 2as
v^2 = 0 + 2(4)(50)
v^2 = 400
v = under root 400
v = 20 cm/sec
Therefore, the velocity after traveling a distance of 50 cm will be 20 cm/sec.
Now,
t = 10sec
u = 20 cm/sec
a = 4 cm/sec^2
Using s = ut + 1/2at^2
s = 20(10) + 1/2(4)(10^2)
s = 200 + 1/2(4)(100)
s = 200 + 2(100)
s = 200 + 200
s = 400 cm
Total distance = 400 + 50
= 450 cm
Therefore, the distance travelled in 10 seconds after the 50 cm point will be 400 cm.
Hope it helps........
Please mark me as the brainliest!!!!!
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