Solve:

The posistion of a particle moving along a coordinate line is s=sqrt(5+4t), with s in meters and t in seconds.
Find the particle's velocity at t=1 sec.

A) 2/3 m/sec
B) 4/3 m/sec
C) -1/3 m/sec
D) 1/6 m/sec

Thank you!

Differentiate s(t). The derivative, ds/dt, is the speed as a function of t.

ds/dt = [(1/2)/sqrt(5 +4t)]*4
= 2/sqrt(5+4t)
At t = 1, this equals 2/(sqrt9) = 2/3

Well, let's first find the derivative of the given position function to get the velocity function. The derivative of s with respect to t gives us:

s' = d(s)/d(t) = d(sqrt(5+4t))/d(t)

Now, let's apply the chain rule. The derivative of sqrt(u) is 1/(2*sqrt(u)) multiplied by the derivative of u with respect to t. In this case, u = 5+4t, so the derivative becomes:

s' = 1/(2*sqrt(5+4t)) * d(5+4t)/d(t)

Now, d(5+4t)/d(t) is simply 4 since the derivative of 5 with respect to t is 0 and the derivative of 4t with respect to t is 4. Plugging this in, we get:

s' = 1/(2*sqrt(5+4t)) * 4

Simplifying further, we obtain:

s' = 2/sqrt(5+4t)

Now let's substitute t = 1 into the velocity function:

s'(1) = 2/sqrt(5+4(1)) = 2/sqrt(9) = 2/3

Therefore, the particle's velocity at t = 1 second is 2/3 m/sec. So, the answer is option A) 2/3 m/sec.

Just remember, velocity and humor go hand in hand. Keep it fun!

To find the velocity of the particle at t=1 sec, we need to take the derivative of the position function with respect to time.

The position function is given as s = sqrt(5 + 4t).

Taking the derivative of s with respect to t, we get:

ds/dt = d/dt(sqrt(5 + 4t))

Applying the chain rule, we have:

ds/dt = (1/2) * (5 + 4t)^(-1/2) * (4)

Simplifying the expression, we get:

ds/dt = 2(5 + 4t)^(-1/2)

Now we can find the velocity at t=1 sec by substituting t=1 into the derivative expression:

ds/dt = 2(5 + 4(1))^(-1/2)

ds/dt = 2(5 + 4)^(-1/2)

ds/dt = 2(9)^(-1/2)

ds/dt = 2/3 m/sec

Therefore, the particle's velocity at t=1 sec is 2/3 m/sec.

Hence, the correct option is A) 2/3 m/sec.

To find the particle's velocity at t=1 sec, we need to calculate the derivative of the position function with respect to time (t). The derivative of a function represents the rate of change of that function.

Let's start by finding the derivative of the position function, s, with respect to time, t:

s = sqrt(5 + 4t)

To find the derivative, we can use the power rule for differentiation. According to the power rule, the derivative of x^n with respect to x is n*x^(n-1).

In this case, the power rule can be applied to the expression inside the square root:

s = (5 + 4t)^(1/2)

Using the power rule, we differentiate the square root:

ds/dt = (1/2) * (5 + 4t)^(-1/2) * d/dt (5 + 4t)

The derivative of 5 is 0 since it is a constant, and the derivative of 4t is simply 4:

ds/dt = (1/2) * (5 + 4t)^(-1/2) * 4

Simplifying further:

ds/dt = 2 * (5 + 4t)^(-1/2)

Now, to find the particle's velocity at t=1 sec, substitute t=1 into the derivative expression:

ds/dt = 2 * (5 + 4(1))^(-1/2)
ds/dt = 2 * (5 + 4)^(-1/2)
ds/dt = 2 * (9)^(-1/2)
ds/dt = 2/3

Therefore, the particle's velocity at t=1 sec is 2/3 m/sec. Hence, the correct option is A) 2/3 m/sec.