A particle moves with a velocity of v(t) = √t − 2 meter/second on a coordinate line.
Find the displacement and the distance travelled by the particle during the given time
interval 0 ≤ t ≤ 3

To find both the **displacement** and the **distance traveled** by the particle over the time interval \( 0 \leq t \leq 3 \), let's go step by step.

### 1. Displacement:
The displacement of the particle is found by integrating the velocity function over the given time interval. 

The velocity function is \( v(t) = \sqrt{t} - 2 \).

Displacement is given by:

\[
\text{Displacement} = \int_{0}^{3} v(t) \, dt = \int_{0}^{3} (\sqrt{t} - 2) \, dt
\]

We will now compute this integral:

\[
\int_{0}^{3} (\sqrt{t} - 2) \, dt = \int_{0}^{3} t^{1/2} \, dt - \int_{0}^{3} 2 \, dt
\]

Now, solve each integral separately:

1. \( \int t^{1/2} \, dt = \frac{2}{3} t^{3/2} \)
2. \( \int 2 \, dt = 2t \)

Now, apply the limits:

\[
\left[ \frac{2}{3} t^{3/2} \right]_{0}^{3} - \left[ 2t \right]_{0}^{3}
\]

Evaluate each term:

\[
= \left( \frac{2}{3} \times 3^{3/2} - \frac{2}{3} \times 0^{3/2} \right) - \left( 2 \times 3 - 2 \times 0 \right)
\]

Simplify:

\[
= \frac{2}{3} \times 3\sqrt{3} - 6 = 2\sqrt{3} - 6
\]

Therefore, the displacement over the time interval is:

\[
\text{Displacement} = 2\sqrt{3} - 6 \approx -2.535 \, \text{meters}
\]

### 2. Distance Travelled:
To find the distance travelled, we need to integrate the absolute value of the velocity, since distance is always positive, regardless of direction.

First, we need to determine where the velocity function changes sign (if at all), because that would indicate a change in direction. We solve \( v(t) = 0 \) to find when the particle changes direction.

\[
\sqrt{t} - 2 = 0
\]

\[
\sqrt{t} = 2 \quad \Rightarrow \quad t = 4
\]

Since \( t = 4 \) is outside the interval \( [0, 3] \), the velocity function is negative for the entire interval. Hence, the absolute value of the velocity is \( |v(t)| = 2 - \sqrt{t} \) over \( [0, 3] \).

The distance travelled is then:

\[
\text{Distance} = \int_{0}^{3} |v(t)| \, dt = \int_{0}^{3} (2 - \sqrt{t}) \, dt
\]

This is the same integral as before, but with a sign change:

\[
= \left[ 2t \right]_{0}^{3} - \left[ \frac{2}{3} t^{3/2} \right]_{0}^{3}
\]

Evaluating gives:

\[
= \left( 2 \times 3 - 2 \times 0 \right) - \left( \frac{2}{3} \times 3^{3/2} - \frac{2}{3} \times 0^{3/2} \right)
\]

\[
= 6 - 2\sqrt{3}
\]

Thus, the total distance travelled is:

\[
\text{Distance} = 6 - 2\sqrt{3} \approx 2.535 \, \text{meters}
\]

### Summary:
- **Displacement**: \( 2\sqrt{3} - 6 \approx -2.535 \, \text{meters} \)
- **Distance traveled**: \( 6 - 2\sqrt{3} \approx 2.535 \, \text{meters} \)

Would you like further clarification on any of these steps?

Here are five follow-up questions to consider:
1. How would the displacement change if the velocity function were different?
2. Can displacement ever be greater than distance traveled?
3. What happens if the time interval is extended beyond \( t = 3 \)?
4. How does the direction of motion affect the sign of displacement?
5. What if the velocity function had a discontinuity? How would that affect the results?

**Tip:** Remember, displacement can be negative, indicating direction, while distance is always positive!

A particle moves with a velocity of v(t) = √t − 2 meter/second on a coordinate line. Find the displacement and the distance travelled by the particle during the given time interval 0 ≤ t ≤ 3.

Solve a calculus problem involving a particle moving with velocity v(t) = √t − 2. Find both displacement and distance travelled over the interval 0 ≤ t ≤ 3 using integration. Suitable for Grades 11-12 students studying calculus and kinematics.

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