plz answer this question
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since you know
-> v=u+at
v=0+100*10=1000m/min
s1=distance acquired during acceleration=ut+1/2*at^2=0+1/2*100*10^2=5000m
-> s2=distance during constant velocity=ut+1/2*at^2=1000*20+0=20000m
since a=0 for constant velocity.
-> during retardation.
since v=u+at
finally car comes to rest so v=0
so 0=1000+a*5
then a=-200m/min^2
s3=(v^2-u^2)/2a =0-1000^2/(-2*200)=2500m
so total displacement=s1+s2+s3=5000+20000+2500
=27500m
i hope you will be able to draw the curve.
by the way if you can't then ask for it i will send you.
-> v=u+at
v=0+100*10=1000m/min
s1=distance acquired during acceleration=ut+1/2*at^2=0+1/2*100*10^2=5000m
-> s2=distance during constant velocity=ut+1/2*at^2=1000*20+0=20000m
since a=0 for constant velocity.
-> during retardation.
since v=u+at
finally car comes to rest so v=0
so 0=1000+a*5
then a=-200m/min^2
s3=(v^2-u^2)/2a =0-1000^2/(-2*200)=2500m
so total displacement=s1+s2+s3=5000+20000+2500
=27500m
i hope you will be able to draw the curve.
by the way if you can't then ask for it i will send you.
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