Solution:
Given:
- Initial velocity: \( u = 20 \, \text{m/s} \)
- Time: \( t = 2 \, \text{seconds} \)
- Distance to the barrier: \( s = 20 \, \text{m} \)
Using the equation of motion:
\[
s = ut + \frac{1}{2} a t^2
\]
\[
20 = 20 \times 2 + \frac{1}{2} a \times 2^2
\]
\[
20 = 40 + 2a
\]
\[
2a = -20
\]
\[
a = -10 \, \text{m/s}^2
\]
Thus, the magnitude of the deceleration is 10 m/s².
Leave a Reply