a body is moved with a speed of 36 km per hour is brought to rest in 10 seconds what is the negative acceleration and the distance travelled by the body before coming to rest
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Answered by
1
u(initial velocity)=10km/s
v(final velocity)=0km/s(since distance travelled by the body come to rest
as you know the acceleration is rate of change of velocity.
v=u+at
(-1)=am/s²
and s
a =50m
I hope it works.
v(final velocity)=0km/s(since distance travelled by the body come to rest
as you know the acceleration is rate of change of velocity.
v=u+at
(-1)=am/s²
and s
a =50m
I hope it works.
Answered by
1
Hii Friend
Good evening
Lets solve This problem
Q.=Easy
Solution:
Given
Initial velocity=36km/h=(36×15/8)=67.5m/s
Final velocity(v)=0 (Brakes applied and body comes in rest)
Time taken=10sec
Use 1st equation of motion
V=u+at
=>0=67.5+a×10
=>-6.75=a
To find Distance(s) use 3rd equation of motion
2as=v2-u2
2×(-6.75)s=0-(-67.5)sq
2×(-6.75)s=-4556.25
s=-4556.25/2×(-6.75)=337.5m
Answer=Negative acceleration=(-6.75)m/s2
Distance =337.5m
Answered by Mr.BEST
See YOU soon
TILL then stay happy stay best.
Good evening
Lets solve This problem
Q.=Easy
Solution:
Given
Initial velocity=36km/h=(36×15/8)=67.5m/s
Final velocity(v)=0 (Brakes applied and body comes in rest)
Time taken=10sec
Use 1st equation of motion
V=u+at
=>0=67.5+a×10
=>-6.75=a
To find Distance(s) use 3rd equation of motion
2as=v2-u2
2×(-6.75)s=0-(-67.5)sq
2×(-6.75)s=-4556.25
s=-4556.25/2×(-6.75)=337.5m
Answer=Negative acceleration=(-6.75)m/s2
Distance =337.5m
Answered by Mr.BEST
See YOU soon
TILL then stay happy stay best.
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