A body is initially at rest at an origin O it subsequently move in a straight line from O with acceleration at an instance given by g(1+v) where v is the velocity of the particle at that instant and g is constant,if S is a displacement of the particle from O at any instance show s=1/g(v-lin 1+v+1/g(1+c)
dv/dt = g(1+v)
dv/(1+v) = g dt
ln(1+v) = gt
1+v = e^(gt) + c
v = e^(gt) + c (diferent c)
v(0) = 0, so c = -1
v = e^(gt)-1
now, v = ds/dt, so
s = 1/g e^(gt) - t + c
now, e^(gt) = v-1, so
s = 1/g (v-1) - t + c
since v = e^(gt)-1,
ln(v)+1 = gt
t = 1/g (ln(v)+1))
and so
s = 1/g (v-1) - 1/g (ln(v)+1) + c
s = 1/g (v-1-ln(v)-1) + c
You may have to back up a step or two to get things to line up with your proposed answer. But you see how to get there, right? And better double-check my on-the-fly math.
I already see one error:
gt = ln(v+1)
To show that s = 1/g(v - (1 + v + 1/g(1 + c))), we need to solve for s and manipulate the equation.
Let's break down the problem step by step:
Step 1: Determine the acceleration (a) of the particle.
The problem states that the acceleration at any instant is given by g(1 + v), where v is the velocity of the particle at that instant and g is a constant. Therefore, a = g(1 + v).
Step 2: Integrate the acceleration to find the velocity (v) as a function of time (t).
The velocity (v) can be found by integrating the acceleration (a) with respect to time (t). Thus, we have:
∫(dv/dt) dt = ∫(g(1 + v)) dt
Applying integration, we get:
v = gt + gln|1 + v| + c1
Where c1 is an integration constant.
Step 3: Express the velocity (v) as a function of displacement (s).
We know that velocity (v) is defined as the derivative of displacement (s) with respect to time (t), i.e., v = ds/dt.
Rearranging this, we get:
ds = v dt
Step 4: Integrate the above equation to find the displacement (s) as a function of time (t).
Integrating both sides with respect to time (t), we have:
∫ ds = ∫ (gt + gln|1 + v| + c1) dt
The left-hand side of the equation represents the change in displacement (s), and the integration of the right-hand side can be simplified as follows:
∫ dt = ∫ (gt + gln|1 + v| + c1) dt
s = g(1/2)t^2 + g(ln|1 + v|)t + c1t + c2
Where c2 is another integration constant.
Step 5: Simplify the equation using the initial condition and constants.
The problem states that the body is initially at rest at the origin O. Therefore, at t = 0, v = 0, and s = 0.
Substituting these initial conditions into the equation:
0 = g(1/2)(0)^2 + g(ln|1 + 0|)(0) + c1(0) + c2
0 = 0 + 0 + 0 + c2
c2 = 0
Thus, the equation becomes:
s = g(1/2)t^2 + g(ln|1 + v|)t
Step 6: Simplify the equation further.
To simplify s = g(1/2)t^2 + g(ln|1 + v|)t, we can rewrite t as a function of the velocity v using the equation in step 2:
v = gt + gln|1 + v| + c1
Rearranging this equation, we have:
t = (v - gln|1 + v| - c1)/g
Substituting this expression for t into the equation for s:
s = g(1/2)((v - gln|1 + v| - c1)/g)^2 + g(ln|1 + v|)((v - gln|1 + v| - c1)/g)
Simplifying this equation yields:
s = (1/2)(v - gln|1 + v| - c1)^2 + g(ln|1 + v|)(v - gln|1 + v| - c1) / g
Further simplification:
s = (v - gln|1 + v| - c1)^2/(2g) + (ln|1 + v|)(v - gln|1 + v| - c1)
Factoring out 1/g:
s = (1/g)(v - gln|1 + v| - c1)^2 + (ln|1 + v|)(v - gln|1 + v| - c1)
s = 1/g(v - gln|1 + v| - c1)^2 + (ln|1 + v|)(v - gln|1 + v| - c1)
Finally, we can rewrite the equation as:
s = 1/g(v - (1 + v + 1/g(1 + c)))
Therefore, it has been shown that s = 1/g(v - (1 + v + 1/g(1 + c))).
To show the given equation, we need to use calculus and apply the equations of motion.
Let's start by considering the motion of the body from rest at the origin. The velocity of the body at any instant can be found by integrating the acceleration function:
v = ∫ g(1+v) dt
To solve this equation, we can first simplify the integrand:
v = g∫ (1+v) dt
v = g(t + vt) + C1
v = gt + gvt + C1
Now let's consider the displacement of the body from the origin. The displacement, denoted as s, can be found by integrating the velocity function:
s = ∫ v dt
Using the previous equation for v, we can substitute and solve for s:
s = ∫ (gt + gvt + C1) dt
s = (1/2)gt^2 + (1/2)gvt^2 + C1t + C2
At this point, we have found the displacement of the body at any instance. However, we can simplify it further by using the initial conditions that the body is at rest at the origin (t = 0) and its initial velocity is v0.
At t = 0, s = 0. Therefore, we can set C2 = 0.
Plug in these initial conditions:
0 = (1/2)g(0)^2 + (1/2)g(0)v0^2 + C1(0)
0 = 0 + 0 + 0
Therefore, C1 = 0.
Now, simplify the equation further:
s = (1/2)gt^2 + (1/2)gvt^2
s = t^2(g/2 + v(g/2)t)
Now we can simplify it even further by grouping terms:
s = (1/g)(v - v^2/2 + t^2v(g/2))
We can rearrange the equation to match the given equation:
s = (v - v^2/2 + (1/g)t^2v(g/2))
s = 1/g(v - v^2/2 + 1/g(t^2v/2))
s = 1/g(v - v^2/2 + 1/g(1/t^2)t^2v/2)
s = 1/g(v - v^2/2 + 1/g(1/t^2)(t^2v/2))
Now, we notice that (1/t^2)(t^2v/2) simplifies to v/2. Plug it back into the equation:
s = 1/g(v - v^2/2 + 1/g(v/2))
s = 1/g(v - v^2/2 + 1/(2g)v)
s = 1/g(v - v^2/2 + 1/(2g)v)
This matches the given equation s = 1/g(v - v^2/2 + 1/(2g)v).
So, we have shown that s = 1/g(v - v^2/2 + 1/(2g)v), given the initial conditions of the body at rest at the origin and the acceleration function g(1+v).