a body leaving a certain point"0" moves with a constant acceleration. At the end of the fifth second it's velocity is 1.5 m/s. At the end of the sixth second the body stops and then begins to move backwards. Find the distance traversed by the body before it stops. Determine the velocity with which the body returns to point "0"? (27m,-9m/s)
Answers
Answer:
v = u + at
1.5 m/s = u + 5a ------------------ (1)
0 m/s = 1.5 m/s + a
⇒ a = − 1.5 m/s2
Substituting the value of a in Eq. (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 initial velocity u = 0, distance covered S = 27 m.
To determine the velocity with which the body returns to point o, we need the time taken or the acceleration in the backward direction. The we can substitute in v = u + at and get the answer.
Explanation:
hope this helps you .