The velocity of a particle is given by v=2x+1 cms^-1. If the particle is initially at the origin, find what the displacement will be when the acceleration is 5 cm s^-2.
Answer 0.75cm
huh? a = dv/dt = 2
the acceleration is a constant 2 cm/s^2, so it will never be 5.
Thats what the Q says though. I dont know how acc can be 5
maybe it means when v=5.
In that case, t=2
so s = x^2+x = 2^2+2 = 6
Hmmm. Definitely not 0.75
so, when is x^2+x = 3/4?
when x = 1/2
make of that what you will
To find the displacement of the particle when the acceleration is 5 cm/s^2, we need to determine the position function by integrating the velocity function.
Given that the velocity function v = 2x + 1 cm/s, we can integrate it with respect to time to get the position function.
∫v dt = ∫(2x + 1) dt
To integrate the function, we need to know that the derivative of position with respect to time is velocity. Considering that dx/dt = v, we can rewrite the equation as:
∫dx = ∫(2x + 1) dt
Now, we integrate both sides of the equation:
x = ∫(2x + 1) dt
x = ∫2x dt + ∫1 dt
x = 2∫x dt + t + C
Here, C is the constant of integration and t represents time.
Now, we'll solve for x:
x - 2∫x dt = t + C
To solve the integral on the left side, we can differentiate both sides with respect to t:
dx/dt - 2x = 1
This is a first-order linear differential equation. We can solve it using an integrating factor. Multiplying the equation by e^(-2t):
e^(-2t) * dx/dt - 2e^(-2t) * x = e^(-2t)
Now, we'll use the product rule on the left side:
(d/dt)(e^(-2t) * x) = e^(-2t)
Integrating both sides with respect to t:
∫(d/dt)(e^(-2t) * x) dt = ∫e^(-2t) dt
e^(-2t) * x = ∫e^(-2t) dt
Integrating the right side gives us:
e^(-2t) * x = -1/2 * e^(-2t) + C'
Where C' is the constant of integration.
Finally, we solve for x:
x = -1/2 + C' * e^(2t)
Since the particle is initially at the origin (x = 0 when t = 0), we can determine the value of the constant C' by substituting these values into the equation:
0 = -1/2 + C' * e^(2*0)
0 = -1/2 + C'
Solving for C', we find C' = 1/2.
Now, we can substitute this value back into our equation for x:
x = -1/2 + (1/2) * e^(2t)
To find the displacement when the acceleration is 5 cm/s^2, we differentiate the position function twice with respect to time.
d^2x/dt^2 = 4e^(2t)
Setting the acceleration equal to 5 cm/s^2:
4e^(2t) = 5
Solving for t:
e^(2t) = 5/4
2t = ln(5/4)
t = 0.25 * ln(5/4)
Substituting this value for t back into the position function:
x = -1/2 + (1/2) * e^(2 * 0.25 * ln(5/4))
Simplifying:
x = -1/2 + (1/2) * (5/4)^2
x = -1/2 + (5/8)
x = 4/8 - 1/2
x = 2/8
x = 0.25 cm
Therefore, the displacement of the particle when the acceleration is 5 cm/s^2 is 0.25 cm.