A particle with velocity $$ v(t)=3 \sqrt{t} $$, where $$ t $$ is time in seconds, moves in a straight line. How far does the particle move from $$ t=1 $$ to $$ t=4 $$ seconds?

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We are given the velocity function
$$
v(t)=3\sqrt{t}.
$$
The distance traveled from $$ t=1 $$ to $$ t=4 $$ seconds is given by the integral
$$
\text{Distance} = \int_{1}^{4} v(t)\,dt = \int_{1}^{4} 3\sqrt{t}\, dt.
$$

**Step 1: Express the integrand in power form**

Recall that $$\sqrt{t}=t^{1/2}$$, so the integrand becomes:
$$
3\sqrt{t} = 3t^{1/2}.
$$

**Step 2: Set up the integral**

We have:
$$
\int_{1}^{4} 3t^{1/2}\, dt.
$$

**Step 3: Factor out the constant**

Pulling 3 out of the integral gives:
$$
3\int_{1}^{4} t^{1/2}\, dt.
$$

**Step 4: Integrate $$ t^{1/2} $$**

The antiderivative of $$ t^{1/2} $$ is:
$$
\int t^{1/2}\, dt = \frac{t^{3/2}}{\frac{3}{2}} = \frac{2}{3} t^{3/2}.
$$

**Step 5: Apply the limits of integration**

Substitute the antiderivative back into the integral:
$$
3\left[\frac{2}{3}t^{3/2}\right]_{t=1}^{t=4}.
$$

The constant factors $$3$$ and $$\frac{2}{3}$$ simplify:
$$
3 \times \frac{2}{3} = 2.
$$
Thus, the expression becomes:
$$
2\left[t^{3/2}\right]_{1}^{4} = 2\left(4^{3/2} - 1^{3/2}\right).
$$

**Step 6: Evaluate the expression**

First, compute $$ 4^{3/2} $$. This is equivalent to:
$$
4^{3/2} = \left(4^{1/2}\right)^3 = (2)^3 = 8.
$$
Next, compute $$ 1^{3/2} $$:
$$
1^{3/2} = 1.
$$
Substitute these values:
$$
2(8 - 1) = 2 \times 7 = 14.
$$

Thus, the particle moves a distance of $$ 14 $$ units from $$ t=1 $$ to $$ t=4 $$ seconds.
The particle moves 14 units from $$ t=1 $$ to $$ t=4 $$ seconds.

A particle with velocity \( v(t)=3 \sqrt{t} \), where \( t \) is time in seconds, moves in a straight line. How far does the particle move from \( t=1 \) to \( t=4 \) seconds?

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