Question - Calculating Final Velocity of a Bicycle Under Constant Acceleration
Solution:
v_{final} = v_{initial} + a * t
v_{final} = 5 m/s + (2 m/s^2 * 3 s)
v_{final} = 5 m/s + 6 m/s
v_{final} = 11 m/s
Tutor Solution:
Let's begin by identifying the given information in the problem and the equation that relates them. The problem provides us with the initial velocity \( v_i \), the acceleration \( a \), and the time \( t \) during which the acceleration occurs, and asks us to find the final velocity \( v_f \).
The known quantities are:
- Initial velocity, \( v_i = 5 \, \text{m/s} \)
- Acceleration, \( a = 2 \, \text{m/s}^2 \)
- Time, \( t = 3 \, \text{s} \)
The final velocity can be found using the kinematic equation which relates velocity, acceleration, and time:
\[ v_f = v_i + a \cdot t \]Now we can substitute the known values into the equation:
\[ v_f = 5 \, \text{m/s} + (2 \, \text{m/s}^2)(3 \, \text{s}) \]Performing the multiplication to determine the increase in velocity due to acceleration over time:
\[ v_f = 5 \, \text{m/s} + 6 \, \text{m/s} \]Adding the initial velocity to the change in velocity to find the final velocity:
\[ v_f = 11 \, \text{m/s} \]Therefore, the final velocity of the bicycle after accelerating for 3 seconds is \( 11 \, \text{m/s} \).
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