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. (1 point) Responses 112 ms/ 1 12 m s 6 ms/ 6 m s 223√ ms/ 2 2 3 m s 14 ms/ 1 4 m s 15 ms/

Question

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. (1 point) Responses 112 ms/ 1 12 m s 6 ms/ 6 m s 223√ ms/ 2 2 3 m s 14 ms/ 1 4 m s 15 ms/

Answer 1

To find the particle's velocity, we need to find the derivative of the position function with respect to time. Taking the derivative of s = 6 + 3t^2, we get:

v = ds/dt = d/dt (6 + 3t^2) = 0 + 6(2t) = 12t

Now we can find the velocity at t = 10 seconds by plugging in t = 10 into the velocity function:

v(10) = 12(10) = 120 m/s

Therefore, the particle's velocity at t = 10 seconds is 120 m/s.

Answer 2

To find the velocity of the particle at t = 10 seconds, we need to differentiate the position function, s, with respect to time, t.

Taking the derivative, we have:

v = ds/dt

Given that s = 6 + 3√t, we can differentiate using the power rule for derivatives:

v = d (6 + 3√t)/dt
= d(6)/dt + d(3√t)/dt
= 0 + 3(1/2t^(-1/2))
= 3(1/2t^(-1/2))
= (3/2)(1/√t)
= 3/(2√t)

Now we can substitute t = 10 seconds into the velocity equation:

v = 3/(2√10)
= 3/(2√(2^2 * 5))
= 3/(2 * 2√5)
= 3/(4√5)
= 3√5/4

So, the particle's velocity at t = 10 seconds is 3√5/4 m/s.

Answer 3

To find the particle's velocity, we need to take the derivative of the position function with respect to time. The derivative of a function represents the rate of change of that function with respect to the independent variable.

Given that the position function is s = 6 + 3√(t), we can differentiate it to find the velocity function:

ds/dt = d(6 + 3√(t))/dt

The derivative of a constant term (6 in this case) is zero, so we only need to differentiate 3√(t):

ds/dt = 3 * d(√(t))/dt

To differentiate √(t), we can use the chain rule, which states that if we have a function of the form f(g(t)), the derivative is f'(g(t)) * g'(t).

In this case, f(x) = √(x) and g(t) = t, so:

d(√(t))/dt = (1/2√(t)) * dt/dt

Since dt/dt = 1, we obtain:

d(√(t))/dt = (1/2√(t))

Substituting this back into the previous equation, we get:

ds/dt = 3 * (1/2√(t))

Simplifying further:

ds/dt = 3/2√(t)

Now we can find the particle's velocity at t = 10 seconds by plugging in this value into the velocity function:

v(10) = 3/2√(10)

Calculating this value, we get:

v(10) ≈ 3/2√10 ≈ 3/2 * 3.16 ≈ 4.74 m/s

Therefore, the particle's velocity at t = 10 seconds is approximately 4.74 m/s.