A car starts from rest, and accelerates uniformly in a straight line until it reaches a speed of 80.0 km h–1 in 5.0 s. What distance did the car cover when going from 20.0 km h–1 to 70.0 km h–1?

Give your answer to the nearest metre.

To solve this problem, we can break it down into steps using kinematic equations. We are interested in the distance the car covers as it accelerates uniformly from 20.0 km/h to 70.0 km/h.

### Step 1: Convert the velocities from km/h to m/s.
\[
\text{Initial velocity (v}_1\text{)} = 20.0 \, \text{km/h} = \frac{20.0 \times 1000 \, \text{m}}{3600 \, \text{s}} = 5.56 \, \text{m/s}
\]
\[
\text{Final velocity (v}_2\text{)} = 70.0 \, \text{km/h} = \frac{70.0 \times 1000 \, \text{m}}{3600 \, \text{s}} = 19.44 \, \text{m/s}
\]

### Step 2: Determine the acceleration.
The car starts from rest and reaches 80.0 km/h in 5.0 seconds, so we can find the acceleration.
\[
\text{Final velocity (v)} = 80.0 \, \text{km/h} = \frac{80.0 \times 1000 \, \text{m}}{3600 \, \text{s}} = 22.22 \, \text{m/s}
\]
\[
\text{Acceleration (a)} = \frac{\text{Final velocity} - \text{Initial velocity}}{\text{Time}} = \frac{22.22 \, \text{m/s} - 0 \, \text{m/s}}{5.0 \, \text{s}} = 4.44 \, \text{m/s}^2
\]

### Step 3: Use the kinematic equation to find the distance.
We use the equation:
\[
v_2^2 = v_1^2 + 2ad
\]
where \( v_2 \) is the final velocity, \( v_1 \) is the initial velocity, \( a \) is the acceleration, and \( d \) is the distance covered.

Rearranging to solve for \( d \):
\[
d = \frac{v_2^2 - v_1^2}{2a}
\]
Substituting the known values:
\[
d = \frac{(19.44 \, \text{m/s})^2 - (5.56 \, \text{m/s})^2}{2 \times 4.44 \, \text{m/s}^2}
\]
\[
d = \frac{377.23 - 30.92}{8.88} \, \text{m}
\]
\[
d = \frac{346.31 \, \text{m}^2/\text{s}^2}{8.88 \, \text{m/s}^2} = 39.00 \, \text{m}
\]

### Final Answer
The car covers approximately **39 metres** when accelerating from 20.0 km/h to 70.0 km/h.

Would you like further details or have any questions?

---

Here are five related questions:
1. How would the distance change if the car's acceleration was doubled?
2. What time would it take for the car to go from 20.0 km/h to 70.0 km/h?
3. What is the significance of uniform acceleration in this problem?
4. How does the distance covered relate to the time if the acceleration is not constant?
5. What is the formula for distance covered in uniformly accelerated motion?

**Tip:** Always check units for consistency when solving kinematic equations, especially when converting between km/h and m/s.

Learn how to calculate the distance a car covers when accelerating uniformly from 20.0 km/h to 70.0 km/h using kinematic equations. This problem involves converting units, determining acceleration, and applying kinematic formulas. Suitable for high school students studying kinematics and uniform acceleration.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation