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A car accelerates from rest a constant rate for first 10 sec and covers a distance of X. It covers distance y in next 10 sec at the same acceleration.
y=3xfirst 10 seconds:v=u + at (u=0)v = 10a ( t= 10)v2 = u2 + 2as100a2 = 2axx = 50a 2nd bodynow u = 10av = 10a + 10a = 20a (v= u + at)v2 = u2 + 2as400a2 = 100a2 + 2aysolving u get y = 150atherefore y = 3x
y=3x
first 10 seconds:
v=u + at (u=0)
v = 10a ( t= 10)
v2 = u2 + 2as
100a2 = 2ax
x = 50a
2nd body
now u = 10a
v = 10a + 10a = 20a (v= u + at)
400a2 = 100a2 + 2ay
solving u get y = 150a
therefore y = 3x
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y=3xfor first 10 secondsv=u + at (and since u=0,started from rest)v = 10a ( till t= 10)v2 = u2 + 2as10a*10a = 2axx = 50a2nd journey case(next 10 sec)(or from t=10sec to t=20sec)now u = 10av = 10a + a*10 = 20a (v= u + at)v2 = u2 + 2as20a*20a= 10a*10a + 2aysolving u get y = 150atherefore y = 3x
for first 10 seconds
v=u + at (and since u=0,started from rest)
v = 10a ( till t= 10)
10a*10a = 2ax
2nd journey case(next 10 sec)(or from t=10sec to t=20sec)
v = 10a + a*10 = 20a (v= u + at)
20a*20a= 10a*10a + 2ay
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