a particle moves along the x-axis at a velocity of v(t)=1 sqrt(t), t>0. at time t=1 , its position is x=4. find the acceleration and position functions for the particle
I suspect a typo. Why write "1 sqrt(t)"?
To find the acceleration function and position function for the particle, let's start by finding the derivative of the velocity function.
Given that the velocity function is v(t) = √t, we can find the acceleration function by taking the derivative. The derivative of v(t) with respect to t (denoted as v'(t) or dv/dt) will give us the acceleration.
Let's differentiate v(t) = √t using the power rule of differentiation:
v'(t) = (1/2) * t^(-1/2)
So, the acceleration function is a(t) = (1/2) * t^(-1/2).
To find the position function, we need to integrate the velocity function. The indefinite integral of v(t) with respect to t (denoted as ∫v(t) dt) will give us the position function.
Let's integrate v(t) = √t:
∫v(t) dt = ∫√t dt
= (2/3) * t^(3/2) + C
Given that the position at time t = 1 is x = 4, we can find the constant C by substituting the values into the position function.
When t = 1, x = 4:
(2/3) * (1)^(3/2) + C = 4
(2/3) + C = 4
To solve for C, subtract 2/3 from both sides:
C = 4 - (2/3)
C = 12/3 - 2/3
C = 10/3
Thus, the position function for the particle is x(t) = (2/3) * t^(3/2) + 10/3.
In summary:
- The acceleration function is a(t) = (1/2) * t^(-1/2).
- The position function is x(t) = (2/3) * t^(3/2) + 10/3.