Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x''(t) is its acceleration.
A particle moves along the x-axis at a velocity of v(t) = 5/√t, t > 0. At time t = 1, its position is x = 11. Find the acceleration and position functions for the particle.
a(t) =
v = 5 t^-.5 = 5/sqrt(t)
a = dv/dt = -2.5 t^-2.5
x = (5/.5)t^.5+c = 10 t^.5 + c
11 = 10 (1)^.5 + c
c = 1
so
x = 10 sqrt(t) + 1
To find the acceleration function, we need to differentiate the velocity function with respect to time (t).
Given v(t) = 5/√t, we can differentiate it to find the acceleration function a(t).
Differentiating v(t) = 5/√t with respect to t, we get:
a(t) = d(v(t))/dt = d(5/√t)/dt
To differentiate 5/√t, we can use the power rule for differentiation:
a(t) = d(5t^(-1/2))/dt = -1/2 * 5t^(-1/2 - 1) = -5/2t^(3/2)
So, the acceleration function is a(t) = -5/2t^(3/2).
Now, to find the position function, we can integrate the velocity function with respect to t.
Given v(t) = 5/√t, we can integrate it to find the position function x(t):
∫ v(t) dt = ∫ (5/√t) dt
Using the power rule for integration, the integral becomes:
x(t) = 2(5√t) + C
where C is the constant of integration.
Given that at time t = 1, x = 11, we can substitute these values into the equation to solve for C:
11 = 2(5√1) + C
11 = 10 + C
C = 1
Therefore, the position function is x(t) = 10√t + 1.
To summarize:
- The acceleration function is a(t) = -5/2t^(3/2).
- The position function is x(t) = 10√t + 1.
To find the acceleration function, we need to take the derivative of the velocity function with respect to time.
Given that the velocity function is v(t) = 5/√t, we can use the power rule of differentiation to find the derivative.
Taking the derivative of v(t), we get:
a(t) = d/dt (5/√t)
Using the power rule, we can rewrite the expression as:
a(t) = 5 * d/dt (t^(-1/2))
Now, applying the power rule again, we can find the derivative of t^(-1/2):
a(t) = 5 * (-1/2) * t^(-3/2)
Simplifying further, we get:
a(t) = -5/2 * t^(-3/2)
So, the acceleration function is a(t) = -5/(2sqrt(t)).
Next, to find the position function, we need to integrate the velocity function with respect to time.
Given that the position function is x(t), we can use the power rule of integration to find the integral.
Integrating v(t), we get:
x(t) = ∫ (5/√t) dt
To integrate this, we use the power rule for integration:
x(t) = 5 * ∫ t^(-1/2) dt
Now, applying the power rule of integration, we get:
x(t) = 5 * (2 * t^(1/2))
Simplifying further, we get:
x(t) = 10 * √t
So, the position function is x(t) = 10√t.
Therefore, the acceleration function is a(t) = -5/(2sqrt(t)) and the position function is x(t) = 10√t.