Question

a body leaving a certain point"o"moves with an a constant acceleration. At the end of the 5th second its velocityis 1.5 m/s at the end of the 6th 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"o"?(27m,-9m/s)

Answer 1

1.8m/s for5s v=1.5m/s. Therefore for6s=1.5/5*6=1.8m/s

Answer 2

more_horiz A body leaving a certain point "o" moves with anconstantacceleration. At the end of the fifthseconditsvelocityis 1.5metersper secon..
A body leaving a certain point "o" moves with an constant acceleration. At the end of the fifth second its velocity is 1.5 meters per second. At the end of sixth second the body stops and then begins to move backward. Find the distance traversed by the body before it stops. Determine the velocity with which the body returns to point "o"?
v = u + at
1.5 m/s = u + 5a ------------------ (1)
0 m/s = 1.5 m/s + a
⇒ a = − 1.5 m/s
Substituting the value of a in Eq. (1),
1.5 m/s = u + 5 x ( − 1.5 m/s )
⇒ u = 9 m/s
Distance travelled in 6 s (i.e. before it stops) S = ut + ½ at
= 9 x 6 + ½ ( − 1.5 m/s ) 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.