Asked by parul
a body moves with a uniform velocity of 2m/sec for 5 sec then velocity was uniformly accelerated and a velocity of 10m/sec was achieved in next 5 sec.then brakes were applied uniformly and body comes to rest in next 10 sec.calculate
a)acceleration and retardation produced
b)distance moved by the body with uniform velocity
c)different distances moved by the body with non-uniform velocity i.e.when acceleration was active
d)average velocity in whole journey
a)acceleration and retardation produced
b)distance moved by the body with uniform velocity
c)different distances moved by the body with non-uniform velocity i.e.when acceleration was active
d)average velocity in whole journey
Answers
Answered by
Elena
s1=v1•t1 =2•5 =10 m.
v2=v1+a1•t2 .
a1 =(v2-v1)/t2 =( 10-5)/5 = 1m/s².
s2 = v1•t2 + a1•t2²/2 =5•5+1•25/2 =37.5 m.
0 =v2+a2•t3,
a2 =(0-v2)/t3 = -10/10 =-1 m/s².
s3 = v2•t3 – a2•t3²/2 =10•10-1•100/2 =50 m
v(ave_ =s/t =(s1+s2+s3)/(t1+t2+t3) =
(10+37.5+50)/(5+5+10) = 97.5/20 =4.875 m/s
v2=v1+a1•t2 .
a1 =(v2-v1)/t2 =( 10-5)/5 = 1m/s².
s2 = v1•t2 + a1•t2²/2 =5•5+1•25/2 =37.5 m.
0 =v2+a2•t3,
a2 =(0-v2)/t3 = -10/10 =-1 m/s².
s3 = v2•t3 – a2•t3²/2 =10•10-1•100/2 =50 m
v(ave_ =s/t =(s1+s2+s3)/(t1+t2+t3) =
(10+37.5+50)/(5+5+10) = 97.5/20 =4.875 m/s
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