A car increases its velocity from rest at 2.39 m/s2 and drives 7.57 m. Calculate how long it took.

To calculate the time it took for the car to drive 7.57 meters while accelerating from rest at \( 2.39 \, \text{m/s}^2 \), we can use the following kinematic equation:

\[
d = v_0 t + \frac{1}{2} a t^2
\]

Where:
- \( d = 7.57 \, \text{m} \) (the distance traveled),
- \( v_0 = 0 \, \text{m/s} \) (initial velocity since the car starts from rest),
- \( a = 2.39 \, \text{m/s}^2 \) (acceleration),
- \( t \) is the time (what we're solving for).

Since the initial velocity \( v_0 = 0 \), the equation simplifies to:

\[
d = \frac{1}{2} a t^2
\]

Solving for \( t \):

\[
t^2 = \frac{2d}{a}
\]

\[
t = \sqrt{\frac{2d}{a}}
\]

Now, substitute the known values:

\[
t = \sqrt{\frac{2 \times 7.57}{2.39}}
\]

Let's calculate the time.The time it took for the car to drive 7.57 meters while accelerating from rest is approximately \( 2.52 \, \text{seconds} \).

Would you like further clarification or details on this solution?

Here are 5 related questions to consider:
1. What is the final velocity of the car after 2.52 seconds?
2. How far would the car travel if the acceleration was doubled?
3. What would the time be if the car traveled a distance of 15 meters under the same conditions?
4. How would the time change if the car started with an initial velocity of 5 m/s?
5. What would the car's acceleration need to be if it traveled 7.57 meters in 3 seconds?

**Tip:** When using kinematic equations, be mindful of the units to ensure consistency in your calculations.

A car increases its velocity from rest at 2.39 m/s² and drives 7.57 m. Calculate how long it took.

Learn how to calculate the time it takes for a car to travel 7.57 meters while accelerating from rest at 2.39 m/s². This problem involves key concepts from kinematics and the use of kinematic equations, suitable for students in Grades 9-12.

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