A car accelerates from rest at constant rate for t seconds and covers a distance x .The distance covered by it in next t second will be ?
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140
u = 0
t= t sec
s = x m
as , s = ut + (1/2) at²
⇒x = 0 + (1/2) at²
⇒a = 2x/t² m/s²
and, v = u +at
= 0 +(2x/t²)×t
= 2x/t
now, for next t sec,
u = 2x/t m/s
s = ut + (1/2) at²
= (2x/t)×t + (1/2)×(2x/t²)×t²
=2x +x = 3x
the distance covered in next t sec will be 3x
t= t sec
s = x m
as , s = ut + (1/2) at²
⇒x = 0 + (1/2) at²
⇒a = 2x/t² m/s²
and, v = u +at
= 0 +(2x/t²)×t
= 2x/t
now, for next t sec,
u = 2x/t m/s
s = ut + (1/2) at²
= (2x/t)×t + (1/2)×(2x/t²)×t²
=2x +x = 3x
the distance covered in next t sec will be 3x
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Explanation:
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