the position of a particle moving along a coordinate line is s=6+3t−−−−−√ , with s in meters and t in seconds. Find the particle’s velocity at t = 10 seconds.
To find the velocity of the particle, we need to take the derivative of the position function with respect to time.
Given: s = 6 + 3√t
Taking the derivative of the position function, we get:
v = d/dt (6 + 3√t)
To find the derivative of √t, we can use the chain rule, which states that for a function f(g(t)), the derivative is given by f'(g(t)) * g'(t).
In this case, f(u) = √u, and g(t) = t. So, the derivative of √t is given by:
d/dt (√t) = (1/2√t) * (1) = 1/(2√t)
Substituting this in the velocity equation, we get:
v = d/dt (6 + 3√t)
= 0 + 3 * (1/(2√t))
= 3/(2√t)
Now we can find the particle's velocity at t = 10 seconds:
v(10) = 3/(2√10)
= (3/2) √10
Therefore, the particle's velocity at t = 10 seconds is (3/2) √10 m/s.
To find the particle's velocity at t = 10 seconds, we need to take the derivative of the position function with respect to time. The derivative of the position function will give us the rate at which the position is changing, which is the velocity. Let's find the derivative of the position function and evaluate it at t = 10 seconds.
Given: s = 6 + 3√t
Take the derivative of s with respect to t:
ds/dt = d(6 + 3√t)/dt
To differentiate 6, we get 0 because it's a constant.
To differentiate 3√t, we use the power rule for differentiation. The derivative of √t with respect to t is (1/2) * t^(-1/2). Multiplying it by 3 gives us (3/2) * t^(-1/2).
So, ds/dt = 0 + (3/2) * t^(-1/2)
Evaluate the derivative at t = 10 seconds:
ds/dt = (3/2) * (10)^(-1/2)
Now let's simplify the expression:
ds/dt = (3/2) * (√10)^(-1)
To simplify further, multiply the numerator and denominator by (√10):
ds/dt = (3/2) / (√10)
Therefore, the particle's velocity at t = 10 seconds is (3/2) / (√10) meters per second.
To find the particle's velocity at t = 10 seconds, we need to find the derivative of the position equation with respect to time (t).
Given that the position equation is s = 6 + 3√t, we can differentiate it to find the velocity function.
Differentiating s with respect to t, we get:
v = ds/dt = d/dt (6 + 3√t)
To differentiate the square root term, we use the chain rule. Let u = √t, then du/dt = 1/(2√t).
Differentiating the equation again, we get:
v = d/dt (6 + 3√t)
= d/dt (6 + 3u)
= 3 * du/dt
= 3 * (1/(2√t))
= 3/(2√t)
Now we can substitute t = 10 seconds into the velocity equation:
v(10) = 3/(2√10)
Simplifying this expression further, we get:
v(10) = 3/(2 * √10)
= 3/(2 * √10) * (√10/√10) <-- rationalize the denominator
= 3√10 / 2 * 10
= 3√10 / 20
Therefore, the particle's velocity at t = 10 seconds is 3√10 / 20 m/s.