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An object moving with constant acceleration along a horizontal path covers the distance between two points 60m apart in 6 seconds.its speed as it passes the second point is 15m/s.find (a)the speed at the first point (b)it's acceleration(c)the distance from point where the object was at rest at the first point. |
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Answer» u= 5 m/ s.Explanation:After 6s, it is given the final velocity is 15 m/ s.Therefore,15 = u + 6a2 × a × 60 = 15^2 - u^2120 = a^2+ 36A^2 + 12au - 2^2120 a = 36a^ + 12au120 a = 36a + 12au 10 = 3A + u15 = u + 3a + 3a15 = 10 + 3a a = 5/ 3m / 5^2Using this, we get u = 5m/ s.Hope it's help ! ...........Thank you...... |
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