from equation t equal to root x+3 find velocity and acceleration t equal to 1second
What is x?
x is distance
To find the velocity and acceleration from the equation t = √(x+3), we need to differentiate the equation with respect to time (t).
First, let's differentiate the equation with respect to t considering x as a function of t.
t = √(x+3)
Differentiating both sides with respect to t using the chain rule, we have:
1 = (1/2)(x+3)^(-1/2) * dx/dt
Now, let's solve for dx/dt (velocity):
dx/dt = 2 / (x+3)^(1/2)
To find the velocity at t = 1 second, we need to substitute x = t^2 - 3:
dx/dt = 2 / ((t^2 - 3) + 3)^(1/2)
= 2 / t^(1/2)
= 2 / √t
Now, let's calculate the velocity at t = 1 second:
v = dx/dt = 2/√1 = 2 m/s
To find the acceleration, we need to differentiate the velocity equation with respect to time (t):
dv/dt = d(dx/dt)/dt = d²x/dt²
Differentiating dx/dt = 2/√t using the quotient rule, we get:
dv/dt = d²x/dt² = -1/2 * 2/(t^(3/2))
= -1/t^(3/2)
Now, let's calculate the acceleration at t = 1 second:
a = dv/dt = -1/(1^(3/2))
= -1 m/s²
Therefore, the velocity at t = 1 second is 2 m/s and the acceleration at t = 1 second is -1 m/s².