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Sanjana06
@Sanjana06
April 2021
2
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Solve 1 question as fast as possible
with detailed explanation
Answers & Comments
gudan1
We have used the two equations V is equal to you + 80 and S is equal to UT plus half a t square
0 votes
Thanks 1
Sanjana06
i didn't understand
Sanjana06
ok I've understood
Sanjana06
thanks
adityakjha24
Time ratio =1:8:1
let say rational constant = x
Let it accelerates with say a in time 'x'
then it reaches a velocity 'v'and continues motion for '8x' time
and finally then retards with 'b' in time 'x'.
v = u + at
v=0+ax
v=ax
60=ax
a=60/x
and, 0=v-bx
b=60/x
Distance covered in the time of acceleration,
S = ut + 1/2at^2
S1=0+1/2at^2
S1=1/2ax^2 = 1/2 * 60/x * x^2 = 30x
Distance covered while moving with constant velocity is,
S2= vt =60*8x = 480x
Distance covered in time of retardation,
v^2 = u^2 + at
=> u^2 = v^2 -at
0^2 =v^2 - 2bS3
S3=60^2/(2*60/x)
S3= 30x
so, total distance = 30x+480x+30x = 540x
Total time = x+8x+x =10x
so,
average velocity = total distance/ toatal time
540x/10x = 54km/h
Hope it helps
1 votes
Thanks 1
adityakjha24
thanks sanjana for brainliest :)
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Answers & Comments
let say rational constant = x
Let it accelerates with say a in time 'x'
then it reaches a velocity 'v'and continues motion for '8x' time
and finally then retards with 'b' in time 'x'.
v = u + at
v=0+ax
v=ax
60=ax
a=60/x
and, 0=v-bx
b=60/x
Distance covered in the time of acceleration,
S = ut + 1/2at^2
S1=0+1/2at^2
S1=1/2ax^2 = 1/2 * 60/x * x^2 = 30x
Distance covered while moving with constant velocity is,
S2= vt =60*8x = 480x
Distance covered in time of retardation,
v^2 = u^2 + at
=> u^2 = v^2 -at
0^2 =v^2 - 2bS3
S3=60^2/(2*60/x)
S3= 30x
so, total distance = 30x+480x+30x = 540x
Total time = x+8x+x =10x
so,
average velocity = total distance/ toatal time
540x/10x = 54km/h
Hope it helps