Answer:

Let the acceleration of the particle be a.

For motion between A and B

u= 12 m/s, s = 40 m, t = 4 s

s = ut + (1/2)at

2

⇒ 40 = 12 x 4 + (1/2) x a x (4)

2

a = -1 ms

−2

For motion between A and C

64= 12t + (1/2)(-L)t

2

=> r

2

- 24t +128 = 0

=>(t- 8) (t- 16) = 0

=> t = 8 s, 16 s

A.The particle will be at C twice, at t = 8 s and t = 16 s.

B.Velocity of the particle at C

At t = 8 s. velocity of the particle V = 12 + (-1) x 8

= 4 m/s.

As the acceleration of the particle is negative, displacement = 0

∴ 0= 12 × t + (1/2)(-1)t

2

=> t = 0, 24 s

At t = 0, velocity at A = 12 m/s,

At t = 24 s, velocity at A is 12 + (-1) x 24 = -12 m/s.

D.The particle reverses the direction of motion at D. For the motion between A and D

u = 12 m/s, v = 0, a = -1 ms

−2

. If AD = s, from the equation v

2

=u

2

+2as

(0)

2

=(12)

2

+ 2 x (-1) x s

s = 72 m

E. After 12 s the particle comes to rest momentarily at D after covering a distance of 72 m.

Distance in SUBSEQUENT 3 s

= 0 x 3 + (1/2) x (-1) x (3)

2

= 4.5 m

∴ d = 72 m + 4.5 m = 76.5 .

EXPLANATION:

hope it helps you....❤