Question

a body leaving a certain point "o" moves with a constant acceleration .At the end of the 5th second its velocity is 1.5m/s.At the end of the 6 th second the body stops and begins to move backwards .find the distance traversed by the body before it stops

Answer 1

Member since Feb 20 2013

v = u + at

1.5 m/s = u + 5a ---- (supposing as equation 1)

0 m/s = 1.5 m/s + a

⇒ a = − 1.5 m/s2

Substituting the value of a in Equation. (1),

1.5 m/s = u + 5 x ( − 1.5 m/s2 )

⇒ u = 9 m/s

Distance travelled in 6 s (i.e. before it stops) S = ut + ½ at2

= 9 x 6 + ½ ( − 1.5 m/s2 ) x 6 x 6

= 54 - 27 = 27 m

On the return journey we know that;

(i) initial velocity u = 0,

(ii) distance covered S = 27 m.

Explanation:

With velocity the body returns to point o, we need the time taken or the acceleration in the backward direction.

Substitute in v = u + at and get the answer.