A car moves rectilinearly from station A to B (rest to rest) with an acceleration a varying with distance X from station A as a = C – DX where C and D are constants. The distance between two stations and maximum velocity are
Answers
Answer:
N=2
Explanation:
Given that,
The acceleration is f=a−bx.....(I)
The acceleration is decreasing with increasing x.
Hence, the velocity will be maximum
Where, f= 0,
From equation (I),
a−bx=0
x=
b
a
Now, we know that
f=
dt
dv
f=
dt
dv
×
dx
dx
f=v
dx
dv
Now, we can write,
v
dx
dv
=a−bx
vdv=(a−bx)dx.....(II)
Now, for maximum velocity,
0
∫
v
max
vdv=
0
∫
b
a
(a−bx)dx
[
2
v
2
]
0
v
max
=(ax−b
2
x
2
)
0
b
a
2
v
max
2
=
b
a
2
−
2b
2
ba
2
2
v
max
2
=
2b
2
2ba
2
−ba
2
2
v
max
2
=
2b
a
2
v
max
2
=
b
a
2
v
max
=
b
a
Now, for the maximum distances x between the two stations
On integrating of equation (II)
At A and B, cars at rest so the velocity is zero at the both stations.
Now,
0
∫
0
vdv=
0
∫
x
(a−bx)dx
0=(ax−b
2
x
2
)
0
x
ax−b
2
x
2
=0
a−b
2
x
=0
x=
b
2a
......(III)
And given that,
x=
b
Na
....(IV)
Now, on comparing equation (III) and (IV)
Then,
N=2
Hence, the value of N is 2.