class11 ,physics, one dimension velocity
Answers
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:
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