body moving with speed of 5m/s In 5 sec body comes to rest Find acveleration and distance covered by body
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a=-(5/5)m/s2
=-1m/s2
d=(5×5)m
=25m
=-1m/s2
d=(5×5)m
=25m
Answered by
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Let the final velocity be v = 0
nd the initial velocity be u = 5
then;
a = v -u / t
a = 0- 5 /5
a = -1 m/ s^2
and
distance covered :
S = ut + 1/2 at^2
S = 5×5 + 1/2 ×-1 ×25
S = 25 - 25/2
S = 12.5 m/s
nd the initial velocity be u = 5
then;
a = v -u / t
a = 0- 5 /5
a = -1 m/ s^2
and
distance covered :
S = ut + 1/2 at^2
S = 5×5 + 1/2 ×-1 ×25
S = 25 - 25/2
S = 12.5 m/s
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