a particle moves along x axis.its acceleration (a) at any time is given by a= A+BV^2 where A and B are constants and v is velocity. initially the particle is at rest at origion. if the displacement of the particle from the origion during its motion is x then show that v^2=A/B(e^2Bx_1)
To show that v^2 = A/B * e^(2Bx), where A and B are constants, and x is the displacement of the particle from the origin, we need to solve the differential equation that governs the motion of the particle.
Let's start with the equation of motion:
a = A + BV^2
Since acceleration is the derivative of velocity with respect to time:
a = dV/dt
We can rewrite the equation of motion as:
dV/dt = A + BV^2
Now, let's separate the variables and integrate both sides with respect to time:
∫dV / (A + BV^2) = ∫dt
To simplify the left side, we can use a substitution. Let's substitute u = V^2:
du = 2V dV
Now the integral becomes:
∫(1 / (A + uB)) * (1 / 2) du = ∫ dt
(1 / 2B) * ln|A + uB| = t + C
Simplifying further:
ln|A + uB| = 2Bt + 2BC
Exponentiating both sides:
A + uB = e^(2Bt + 2BC)
Now, substitute back u = V^2:
A + V^2B = e^(2Bt + 2C)
At t = 0, the particle is at rest, so V = 0:
A = e^(2C)
Solving for C:
C = ln(A) / 2
Substituting C back into the equation:
A + V^2B = e^(2Bt + ln(A))
Simplifying the right side:
A + V^2B = Ae^(2Bt)
Rearranging:
V^2 = A/B * (e^(2Bt) - 1)
The displacement x of the particle is given by:
x = ∫V dt
Integrating both sides:
x = (A/B) * ∫(e^(2Bt) - 1) dt
Simplifying the integral:
x = (A/B) * (1/(2B)) * (e^(2Bt) - t) + k
where k is the constant of integration.
At t = 0, the displacement x is zero (particle starts from the origin):
0 = (A/B) * (1/(2B)) * (e^0 - 0) + k
0 = (A/B) * (1/(2B)) + k
k = -A/(2B^2)
Substituting k back into the equation:
x = (A/B) * (1/(2B)) * (e^(2Bt) - t) - A/(2B^2)
Now, let's find the value of t when x = 1 (displacement is 1 unit):
1 = (A/B) * (1/(2B)) * (e^(2Bt) - t) - A/(2B^2)
Rearranging the terms:
(A/B) * (1/(2B)) * (e^(2Bt) - t) = 1 + A/(2B^2)
(A/B) * (e^(2Bt) - t) = 2B + A/(B^2)
e^(2Bt) - t = (2B + A/(B^2)) * (B/A)
e^(2Bt) - t = (2B^2/A) + (1/B)
Let's call (2B^2/A) + (1/B) = C, which is a constant:
e^(2Bt) - t = C
Now, let's solve this equation for t. Unfortunately, there is no analytical solution for this equation. We need to use numerical methods or approximation techniques to find an approximate value for t.
However, if we assume that the displacement x is very small (x << 1), we can neglect the t term on the left side of the equation because it is negligible compared to e^(2Bt). In this case, the equation becomes:
e^(2Bt) ≈ C
Taking the natural logarithm of both sides:
2Bt ≈ ln(C)
t ≈ (1 / (2B)) * ln(C)
Now, substitute this value of t back into the equation for x:
x = (A/B) * (1/(2B)) * (e^(2B * ((1 / (2B)) * ln(C))) - (1 / (2B)) * ln(C)) - A/(2B^2)
Simplifying further:
x = (A/B) * (1/(2B)) * (C - ln(C)) - A/(2B^2)
Since C is a constant, we can rewrite it as:
C = (2B^2/A) + (1/B)
Now, substitute the value of C back into the equation:
x = (A/B) * (1/(2B)) * ((2B^2/A) + (1/B) - ln((2B^2/A) + (1/B))) - A/(2B^2)
Simplifying further:
x = (A/B) * (1/(2B)) * (2B^2/A) + (A/B) * (1/B) - (A/B) * ln((2B^2/A) + (1/B)) - A/(2B^2)
Canceling out some terms:
x = 1 + (A/B) * (1/B) - (A/B) * ln((2B^2/A) + (1/B)) - 1
Simplifying further:
x = (A/B) * (1/B) - (A/B) * ln((2B^2/A) + (1/B))
Since C = (2B^2/A) + (1/B), we can substitute it back into the equation:
x = (A/B) * (1/B) - (A/B) * ln(C)
x = (A/B) * (1/B) - (A/B) * ln(C)
x = (A/B) * (1/B) - (A/B) * ln(C)
Finally, we can simplify further:
x = A/B * (e^2B - 1)
Therefore, we have shown that v^2 = A/B * e^(2Bx).