8. A car accelerates at 2.10m/s2 along a straight road. It passes two marks that are 26.9m apart at times t=4.20s and t=5.00s. What was the car's velocity at t=0?

To solve for the car's velocity at \( t = 0 \), we can use the equations of uniformly accelerated motion. Given dataTo solve for the car's velocity at \( t = 0 \), we can use the equations of uniformly accelerated motion. Given 

- Acceleration \( a = 2.10 \, \text{m/s}^2 \)
- The distance between two points \( \Delta x = 26.9 \, \text{m} \)
- Time at the first mark \( t_1 = 4.20 \, \text{s} \)
- Time at the second mark \( t_2 = 5.00 \, \text{s} \)

### Step 1: Average Velocity Between Two Marks

The average velocity \( v_{\text{avg}} \) between the two times \( t_1 \) and \( t_2 \) can be calculated as:

\[
v_{\text{avg}} = \frac{\Delta x}{\Delta t}
\]

where:

\[
\Delta x = 26.9 \, \text{m}, \quad \Delta t = t_2 - t_1 = 5.00 \, \text{s} - 4.20 \, \text{s} = 0.80 \, \text{s}
\]

So:

\[
v_{\text{avg}} = \frac{26.9}{0.80} = 33.625 \, \text{m/s}
\]

### Step 2: Average Velocity in Terms of Initial and Final Velocity

The average velocity during a time interval is also related to the initial and final velocities by:

\[
v_{\text{avg}} = \frac{v_1 + v_2}{2}
\]

where \( v_1 \) is the velocity at \( t_1 \) and \( v_2 \) is the velocity at \( t_2 \).

### Step 3: Express Final and Initial Velocities

Since the car is accelerating uniformly, we can use the following equation to express the final velocity \( v_2 \) at \( t_2 \):

\[
v_2 = v_1 + a \Delta t
\]

where \( a = 2.10 \, \text{m/s}^2 \) and \( \Delta t = 0.80 \, \text{s} \).

So:

\[
v_2 = v_1 + 2.10 \times 0.80 = v_1 + 1.68 \, \text{m/s}
\]

### Step 4: Set Up Equation and Solve for \( v_1 \)

Now substitute into the equation for average velocity:

\[
33.625 = \frac{v_1 + (v_1 + 1.68)}{2}
\]

Simplify:

\[
33.625 = \frac{2v_1 + 1.68}{2}
\]

Multiply both sides by 2:

\[
67.25 = 2v_1 + 1.68
\]

Solve for \( v_1 \):

\[
2v_1 = 67.25 - 1.68 = 65.57
\]

\[
v_1 = \frac{65.57}{2} = 32.785 \, \text{m/s}
\]

This is the velocity of the car at \( t_1 = 4.20 \, \text{s} \).

### Step 5: Find Velocity at \( t = 0 \)

To find the velocity at \( t = 0 \), we can use the equation of motion:

\[
v_1 = v_0 + a t_1
\]

where \( v_0 \) is the velocity at \( t = 0 \). Rearranging this equation to solve for \( v_0 \):

\[
v_0 = v_1 - a t_1
\]

Substitute the known values:

\[
v_0 = 32.785 - 2.10 \times 4.20
\]

\[
v_0 = 32.785 - 8.82 = 23.965 \, \text{m/s}
\]

Thus, the velocity at \( t = 0 \) is approximately:

\[
v_0 \approx 23.97 \, \text{m/s}
\]

### Final Answer:
The car's velocity at \( t = 0 \) is \( 23.97 \, \text{m/s} \).

---

Would you like further clarification on any part of the solution? Here are some related questions you might consider:

1. How do you derive the equation for uniformly accelerated motion?
2. What would happen to the initial velocity if the acceleration were negative?
3. How does the time difference between marks affect the velocity calculation?
4. Can we find the position of the car at \( t = 0 \) given this data?
5. How can we apply this method to non-constant acceleration?

**Tip:** Remember that average velocity during uniform acceleration is always the midpoint between the initial and final velocities.

A car accelerates at 2.10m/s^2 along a straight road. It passes two marks that are 26.9m apart at times t=4.20s and t=5.00s. What was the car's velocity at t=0?

Learn how to calculate the initial velocity of a car accelerating uniformly at 2.10 m/s^2 along a straight road, given distances and time intervals. This problem involves key kinematic concepts such as average velocity, uniform acceleration, and the equations of motion. Suitable for Grades 10-12.

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