At what displacement the instantaneous velocity become equal to half of its maximum velocity? Plz help me
Depends on the acceleration. If acceleration is constant
vf=at
so when is v=1/2 vf=a*(t/2)
answer, v is one half final velocity at t/2
but displacement is 1/2 a t^2, so because of the square, when time is 1/2 final time, displacement must be 1/4 the final displacement.
To find the displacement at which the instantaneous velocity becomes equal to half of its maximum velocity, we need to use the equation of motion for uniformly accelerated motion.
The equation for velocity as a function of displacement for uniformly accelerated motion is given by:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement
In this case, the maximum velocity would be "v" and half of the maximum velocity would be "v/2".
Therefore, we can write the equation as:
(v/2)^2 = u^2 + 2as
Simplifying the equation:
(v^2)/4 = u^2 + 2as
Rearranging the equation:
2as = (v^2)/4 - u^2
Now, we can substitute the maximum velocity value into the equation. Let the maximum velocity be "v_max".
2as = (v_max^2)/4 - u^2
This equation relates the displacement (s) to the maximum velocity (v_max) and initial velocity (u).
Remember to provide the values of v_max and u to find the displacement (s).
To find the displacement at which the instantaneous velocity becomes equal to half of its maximum velocity, we can use the equation for velocity in terms of displacement.
Assuming the motion is one-dimensional and the acceleration is constant, we can use the equation for velocity:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, we want to find the displacement at which the instantaneous velocity is equal to half of its maximum velocity. Maximum velocity is typically achieved at the midpoint of the motion when acceleration is constant.
Let's assume the displacement when the instantaneous velocity is half of its maximum velocity is "x".
At the midpoint, the initial velocity u = 0, since the object starts from rest. Therefore, the equation becomes:
v = 0 + at/2 ('at' since it is the acceleration at midpoint)
Now, we know the instantaneous velocity at this point is half of the maximum velocity. Let's represent the maximum velocity as 'Vmax':
v = (1/2)Vmax
Substituting the values, we now have:
(1/2)Vmax = (at)/2
Simplifying the equation, we get:
Vmax = at
Now, let's rewrite this equation to solve for time:
t = Vmax / a
Since we assumed the displacement at half of the maximum velocity as 'x', the equation connecting displacement and time is:
x = (1/2)at^2
Rewriting it for t:
t = sqrt(2x / a)
Now, substitute this value of t back into the equation for maximum velocity:
Vmax = a * (sqrt(2x / a)) / a
Simplifying further:
Vmax = sqrt(2ax)
Now, since we want to know the displacement when the instantaneous velocity is half of the maximum velocity (Vmax/2), we can write:
Vmax/2 = sqrt(2ax)
Squaring both sides of the equation:
(Vmax/2)^2 = 2ax
Let's simplify:
Vmax^2 / 4 = 2ax
Rearranging the equation to get displacement:
x = (Vmax^2) / (8a)
Now, you can calculate the value of x by substituting the values of Vmax and a into the equation. Make sure to use consistent units for all the variables.