A car accelerates from 12 m/s to 21 m/s in 6.2 s. What was its acceleration assuming it is constant?
Express your answer using two significant figures.

To find the acceleration, we use the formula:

\[
a = \frac{{v_f - v_i}}{{t}}
\]

where:
- \( v_f \) is the final velocity (21 m/s),
- \( v_i \) is the initial velocity (12 m/s),
- \( t \) is the time (6.2 s),
- \( a \) is the acceleration.

Now, substituting the values:

\[
a = \frac{{21 \, \text{m/s} - 12 \, \text{m/s}}}{{6.2 \, \text{s}}}
\]

\[
a = \frac{{9 \, \text{m/s}}}{{6.2 \, \text{s}}} \approx 1.45 \, \text{m/s}^2
\]

With two significant figures, the acceleration is:

\[
a \approx 1.5 \, \text{m/s}^2
\]

### Final Answer:
The acceleration is \( 1.5 \, \text{m/s}^2 \).

Would you like further details or clarification on any part of this solution?

---

Here are five related questions you might find interesting:
1. What is the distance traveled by the car during this acceleration?
2. If the car continued accelerating at the same rate, what would its velocity be after 10 seconds?
3. What is the average velocity of the car during this acceleration period?
4. How long would it take for the car to go from rest to 21 m/s with the same acceleration?
5. What is the force required to produce this acceleration if the car's mass is 1500 kg?

**Tip**: In constant acceleration problems, always ensure that units are consistent to avoid mistakes!

A car accelerates from 12 m/s to 21 m/s in 6.2 s. What was its acceleration assuming it is constant? Express your answer using two significant figures.

Learn how to calculate the constant acceleration of a car that accelerates from 12 m/s to 21 m/s in 6.2 seconds. This physics problem uses the basic acceleration formula and applies Newton's First Law of Motion. Suitable for students in Grades 9-11.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation