A car accelerates at 0.85 m/s2.  Its velocity changes from 14 m/s to 22 m/s.  How far did
the car travel while it was accelerating?

To solve the problem of finding the distance traveled by the car while it was accelerating, we can use one of the kinematic equations that relates initial velocity, final velocity, acceleration, and distance. The appropriate equation for this scenario is:

\[
v^2 = u^2 + 2as
\]

Where:
- \( v \) is the final velocity (22 m/s),
- \( u \) is the initial velocity (14 m/s),
- \( a \) is the acceleration (0.85 m/s²),
- \( s \) is the distance traveled.

We need to solve for \( s \):

\[
s = \frac{v^2 - u^2}{2a}
\]

Now, substitute the values:

\[
s = \frac{(22)^2 - (14)^2}{2 \times 0.85}
\]

\[
s = \frac{484 - 196}{1.7}
\]

\[
s = \frac{288}{1.7} \approx 169.41 \text{ meters}
\]

So, the car traveled approximately **169.41 meters** while it was accelerating.

Would you like more details or have any questions?

Here are five related questions:
1. How would the distance change if the acceleration were increased to 1.5 m/s²?
2. What would be the time taken for the car to reach the final velocity?
3. If the car continued at the final velocity for 10 more seconds, what would be the additional distance traveled?
4. How would the distance traveled change if the initial velocity was 10 m/s instead of 14 m/s?
5. What would be the car's velocity after traveling 100 meters with the same acceleration?

**Tip:** When dealing with constant acceleration, kinematic equations are powerful tools for solving problems involving distance, velocity, and time.

Learn how to calculate the distance traveled by a car with an acceleration of 0.85 m/s², starting from an initial velocity of 14 m/s and reaching a final velocity of 22 m/s. This problem involves using the equation of motion for constant acceleration and is suitable for high school students.

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