A plane starts from rest and accelerate to 40m/s in 10s. How far did the plane travel during this time?

We can solve this problem using the equations of motion for uniformly accelerated motion. Since the plane starts from rest, its initial velocity \( u = 0 \), its final velocity \( v = 40 \, \text{m/s} \), and the time taken is \( t = 10 \, \text{seconds} \).

### Step-by-step solution:

1. **Acceleration**: We first need to find the acceleration of the plane. Using the first equation of motion:

\[
v = u + at
\]

Substituting the known values:

\[
40 = 0 + a \cdot 10
\]

Solving for \( a \):

\[
a = \frac{40}{10} = 4 \, \text{m/s}^2
\]

2. **Distance traveled**: Now that we know the acceleration, we can use the second equation of motion to find the distance \( s \):

\[
s = ut + \frac{1}{2}at^2
\]

Since the plane starts from rest (\( u = 0 \)):

\[
s = 0 \cdot 10 + \frac{1}{2} \cdot 4 \cdot 10^2
\]

\[
s = 2 \cdot 100 = 200 \, \text{meters}
\]

Thus, the plane traveled **200 meters** during the 10 seconds.

Would you like more details or have any questions about this? 

Here are 5 related questions to explore further:

1. What is the average velocity of the plane during this time?
2. How would the distance change if the plane accelerated at 5 m/s² instead?
3. If the plane took 20 seconds to reach the same velocity, what distance would it cover?
4. How long would it take for the plane to travel 500 meters with the same acceleration?
5. What would happen to the distance traveled if the initial velocity was 10 m/s instead of 0?

**Tip:** In uniformly accelerated motion, the average velocity can be found by taking the average of the initial and final velocities.

A plane starts from rest and accelerates to 40m/s in 10s. How far did the plane travel during this time?

Find out how far a plane travels when accelerating from rest to 40 m/s in 10 seconds. Using equations of motion, this problem explains step-by-step how to calculate the distance covered, suitable for Grades 9-11 students studying kinematics and classical mechanics.

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