from equation t equal to root x+3 find velocity and acceleration t equal to 1second
To find the velocity and acceleration from the equation t = √(x + 3), we need to differentiate the equation with respect to time (t) to find the expressions for velocity (v) and acceleration (a).
Given the equation t = √(x + 3), we need to differentiate it using the chain rule, which states that if we have a function f(g(t)), the derivative of f(g(t)) with respect to t is given by f'(g(t)) * g'(t).
Let's start by finding the derivative of both sides of the equation with respect to t:
d/dt (t) = d/dt (√(x + 3))
The derivative of t with respect to t is simply 1:
1 = d/dt (√(x + 3))
Now, let's differentiate the right side of the equation using the chain rule. The derivative of √(x + 3) with respect to t is:
d/dt (√(x + 3)) = 1 / (2√(x + 3)) * d/dt (x + 3)
Next, we need to find the derivative of (x + 3) with respect to t. However, since we only have an equation involving t, we need to implicitly differentiate with respect to t. So, we treat x as a function of t:
d/dt (x + 3) = d/dt (x) + d/dt (3)
The derivative of x with respect to t is simply dx/dt, and the derivative of a constant is 0:
d/dt (x + 3) = dx/dt + 0
Therefore, the equation becomes:
1 = 1 / (2√(x + 3)) * dx/dt
To find the velocity, we can solve for dx/dt:
dx/dt = 2√(x + 3)
Given t = 1 second, we can substitute t = 1 into the equation to find the value of dx/dt:
dx/dt = 2√(x + 3)
dx/dt = 2√(x + 3)
dx/dt = 2√(1 + 3)
dx/dt = 2√(4)
dx/dt = 2 * 2
dx/dt = 4
Therefore, the velocity (v) at t = 1 second is 4 units per second.
To find the acceleration, we need to find the derivative of velocity (dx/dt) with respect to time (t):
d/dt (dx/dt) = d/dt (4)
Since dx/dt is a constant, its derivative with respect to t is 0:
d/dt (dx/dt) = 0
Therefore, the acceleration (a) at t = 1 second is 0 units per second squared.