A train moves, starting from rest with an acceleration of \( 2 \mathrm{~m} / \mathrm{s}^{2} \) along a straight road. 1. What is its speed (instantaneous) after 6 seconds? Starts from rest means \( v_{0}=0 . \quad v=v_{0}+a t=0+2 \frac{\mathrm{~m}}{\mathrm{~s}^{2}} \times 6 \mathrm{~s}=12 \) 2. What is the distance traveled by the train during that 6 seconds? \( x=x_{0}+v_{0} t+\frac{1}{2} a t^{2} \)

After calculating the instantaneous speed of the train after 6 seconds, we can use the formula to find the distance traveled during that time. With initial position \( x_0 = 0 \), initial speed \( v_0 = 0 \), and acceleration \( a = 2 \, \mathrm{m/s^2} \), the formula \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \) simplifies to: \( x = 0 + 0 \times 6 + \frac{1}{2} \times 2 \times (6)^2 = 36 \, \mathrm{m} \). Therefore, the train traveled 36 meters in those 6 seconds. Want to know what this means in the real world? Imagine a super-fast train picking up speed as it rolls out of the station. In just 6 seconds, while you’re still sipping your coffee, it’s already dashed 36 meters down the tracks! That’s how swiftly acceleration can transform rest into thrilling motion—perfect for commuting or just zooming past the scenery!