A body starts with an initial velocity of 10 M per second and acceleration 5 metre per second find the distance covered by it in 5 seconds
Answers
Answered by
65
Second equation of motion is
S = ut +1÷2at2
Given that : u=10m per second
a =5ms-2
So,
S =10*5+1÷2*5*25
S =50+1÷2*125
S =50+64.5
S =114.5m
S = ut +1÷2at2
Given that : u=10m per second
a =5ms-2
So,
S =10*5+1÷2*5*25
S =50+1÷2*125
S =50+64.5
S =114.5m
Answered by
119
v=u+at
v=10+5(5)
v=10+25=35
then v²-u²=2as
s= (35²-10²)/2(5)= 112.5 m
or s=ut+1/2 at²
= 10(5)+1/2(125)
= 50+62.5
= 112.5 m is distance covered in 5 secs
v=10+5(5)
v=10+25=35
then v²-u²=2as
s= (35²-10²)/2(5)= 112.5 m
or s=ut+1/2 at²
= 10(5)+1/2(125)
= 50+62.5
= 112.5 m is distance covered in 5 secs
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