A car accelerates for 6 seconds from an initial velocity of 10 meters per second. The acceleration is increasing uniformly from 0 to 8 meters per second squared in 6 seconds. The car decelerates at a constant rate of 2 meters per second squared. Compute the total distance the car has traveled from the start after 8 seconds.

To solve this problem, we'll break it down into two parts: the car's motion during acceleration and deceleration.

### Part 1: Acceleration Phase (0 to 6 seconds)

1. **Initial Velocity (\(v_0\))**: 10 m/s
2. **Acceleration (\(a\))**: Increases uniformly from 0 to 8 m/s² over 6 seconds.

Since the acceleration increases uniformly, it is essentially linear, and we can represent it as:

\[
a(t) = \frac{8}{6}t = \frac{4}{3}t
\]

To find the velocity at any time \( t \), integrate the acceleration function:

\[
v(t) = \int a(t) \, dt = \int \frac{4}{3}t \, dt = \frac{2}{3}t^2 + C
\]

Using the initial condition \( v(0) = 10 \), we find:

\[
v(t) = \frac{2}{3}t^2 + 10
\]

To find the distance traveled during this phase, integrate the velocity function:

\[
x(t) = \int v(t) \, dt = \int \left(\frac{2}{3}t^2 + 10\right) \, dt = \frac{2}{9}t^3 + 10t + C
\]

Using the initial condition \( x(0) = 0 \), we have:

\[
x(t) = \frac{2}{9}t^3 + 10t
\]

Now, calculate the distance after 6 seconds:

\[
x(6) = \frac{2}{9}(6)^3 + 10(6) = \frac{2}{9}(216) + 60 = 48 + 60 = 108 \text{ m}
\]

The velocity at 6 seconds is:

\[
v(6) = \frac{2}{3}(6)^2 + 10 = \frac{2}{3}(36) + 10 = 24 + 10 = 34 \text{ m/s}
\]

### Part 2: Deceleration Phase (6 to 8 seconds)

1. **Initial Velocity**: 34 m/s (from the end of the acceleration phase)
2. **Deceleration**: -2 m/s² (constant)

We calculate the velocity at time \( t = 8 \) seconds:

\[
v(8) = v(6) + a(t)(\Delta t) = 34 - 2 \times (8 - 6) = 34 - 4 = 30 \text{ m/s}
\]

Now, calculate the distance covered during deceleration (6 to 8 seconds):

\[
x(t) = v(6) \cdot (t - 6) + \frac{1}{2}(-2)(t - 6)^2
\]

Substituting \( t = 8 \):

\[
x(8) = 34 \cdot (8 - 6) + \frac{1}{2}(-2)(8 - 6)^2 = 34 \cdot 2 - (1 \cdot 4) = 68 - 4 = 64 \text{ m}
\]

### Total Distance Traveled

Add the distances from both phases:

\[
\text{Total Distance} = 108 \text{ m} + 64 \text{ m} = 172 \text{ m}
\]

**Total distance traveled by the car after 8 seconds is 172 meters.**

Would you like more details or have any questions?

Here are some related questions:
1. How would the distance change if the acceleration were not uniformly increasing?
2. What if the deceleration rate was different?
3. How can you find the maximum velocity during acceleration?
4. How does changing initial velocity affect the total distance?
5. What is the total distance if the acceleration continued for the entire 8 seconds?

**Tip:** For uniformly varying acceleration, use integration to determine velocity and position over time.

Learn how to calculate the total distance traveled by a car in 8 seconds, considering uniform acceleration and constant deceleration. This problem covers key concepts in kinematics, including equations of motion and uniform acceleration. Suitable for high school students studying physics and mathematics.

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