A car which is initially at rest accelerates along a straight road at a rate of 3.5 m/s2.  What will be the magnitude of the speed after 15 s?

To find the magnitude of the speed after 15 seconds, we can use the equation of motion:

\[
v = u + at
\]

Where:
- \( v \) is the final velocity (what we need to find),
- \( u \) is the initial velocity (which is 0, as the car starts from rest),
- \( a \) is the acceleration (3.5 m/s²),
- \( t \) is the time (15 seconds).

Substituting the values:

\[
v = 0 + (3.5 \, \text{m/s}^2)(15 \, \text{s})
\]

\[
v = 52.5 \, \text{m/s}
\]

So, the magnitude of the speed after 15 seconds is **52.5 m/s**.

Would you like more details or have any questions?

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Here are 5 related questions to expand your understanding:
1. How far does the car travel in 15 seconds?
2. What would be the speed if the car accelerated for 20 seconds?
3. How long would it take the car to reach a speed of 100 m/s?
4. What is the difference between speed and velocity?
5. How would the final speed change if the car's acceleration were halved?

**Tip:** In problems involving constant acceleration, always start by identifying the known values and selecting the appropriate kinematic equation.

A car which is initially at rest accelerates along a straight road at a rate of 3.5 m/s². What will be the magnitude of the speed after 15 s?

Learn how to calculate the final speed of a car that starts from rest and accelerates at 3.5 m/s² over 15 seconds. This problem demonstrates key kinematic principles, such as constant acceleration and the equation of motion. Suitable for students in Grades 9-11.

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