Question

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.​

Answer 1

Answer:

u= 5 m/ s.

Explanation:

After 6s, it is given the final velocity is 15 m/ s.

Therefore,

15 = u + 6a

2 × a × 60 = 15^2 - u^2

120 = a^2+ 36a^2 + 12au - 2^2

120 a = 36a^ + 12au

120 a = 36a + 12au

10 = 3a + u

15 = u + 3a + 3a

15 = 10 + 3a

a = 5/ 3m / 5^2

Using this, we get u = 5m/ s.

Hope it's help ! ...........

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