We are given the following information:

- \( s = 50 \, \text{m} \) (displacement)
- \( u = 0 \, \text{ms}^{-1} \) (initial velocity)
- \( v = 20 \, \text{ms}^{-1} \) (final velocity)

We need to find the acceleration \( a \) and the time \( t \). We can use the following kinematic equations:

1. **To find the acceleration** \( a \), use the equation:
   \[
   v^2 = u^2 + 2as
   \]
   Rearranging for \( a \):
   \[
   a = \frac{v^2 - u^2}{2s}
   \]
   Substituting the values:
   \[
   a = \frac{(20)^2 - (0)^2}{2 \times 50} = \frac{400}{100} = 4 \, \text{ms}^{-2}
   \]

2. **To find the time** \( t \), use the equation:
   \[
   v = u + at
   \]
   Rearranging for \( t \):
   \[
   t = \frac{v - u}{a}
   \]
   Substituting the values:
   \[
   t = \frac{20 - 0}{4} = 5 \, \text{seconds}
   \]

Thus, the acceleration is \( a = 4 \, \text{ms}^{-2} \) and the time taken is \( t = 5 \, \text{seconds} \).

Would you like more details or have any questions?

Here are 5 related questions:
1. How would the acceleration change if the displacement were doubled?
2. What is the meaning of constant acceleration in this context?
3. How can we calculate the distance if we only know \( v \), \( t \), and \( a \)?
4. What happens to the time taken if the final velocity is doubled?
5. How do you determine if motion is uniformly accelerated?

**Tip:** When solving kinematic equations, always identify the known variables first to decide which formula fits best.

Learn how to find acceleration and time using kinematic equations with given displacement (50m), initial velocity (0 ms^-1), and final velocity (20 ms^-1). This problem covers key physics concepts related to motion and uniformly accelerated motion. Suitable for high school students in Grades 9-12.

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