A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms-1 to 20 ms-1 while passing through a distance 135m in t second. The value of t is (2) 10 (1) 9 (4) 12 (3) 1.8

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Solution

$u = 10 m / s, v = 20 m / s, s = 135 m, v^{2} = u^{2} + 2 a s 400 = 100 + 2 a * 135 a = \frac{400 - 100}{270} = 1.11 t = \frac{20 - 10}{1.11} = 9 s e c . R e g a r d s$

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A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms-1 to 20 ms-1 while passing through a distance 135m in t second. The value of t is (2) 10 (1) 9 (4) 12 (3) 1.8

u=10m/s,v=20m/s, s=135 m , v2=u2+2as400=100+2a*135a=400-100270=1.11t=20-101.11=9 sec.Regards

$u = 10 m / s, v = 20 m / s, s = 135 m, v^{2} = u^{2} + 2 a s 400 = 100 + 2 a * 135 a = \frac{400 - 100}{270} = 1.11 t = \frac{20 - 10}{1.11} = 9 s e c . R e g a r d s$