John is riding his bicycle along a curved path in a park. The path has a radius of 10 meters. He starts from rest and accelerates uniformly at 2 m/s² for a duration of 5 seconds. After that, he maintains a constant speed for 10 seconds before coming to a stop.
a) Calculate John's initial velocity when he starts accelerating.
b) During the 5-second period of acceleration, how far does John travel along the curved path?
a) Calculate John's initial velocity when he starts accelerating.
b) During the 5-second period of acceleration, how far does John travel along the curved path?
Add an Answer
More Questions From This User See All
Answers & Comments
Answer:
Explanation:
a) To calculate John's initial velocity when he starts accelerating, we can use the kinematic equation:
\[v_f = v_i + at\]
Where:
- \(v_f\) is the final velocity (which will be the velocity after 5 seconds)
- \(v_i\) is the initial velocity (which we want to find)
- \(a\) is the acceleration (2 m/s²)
- \(t\) is the time (5 seconds)
Given that \(v_f = 0\) (he starts from rest), \(a = 2 \, \text{m/s}^2\), and \(t = 5 \, \text{seconds}\), we can solve for \(v_i\):
\[0 = v_i + (2 \, \text{m/s}^2) \cdot (5 \, \text{seconds})\]
\[v_i = -10 \, \text{m/s}\]
So, John's initial velocity when he starts accelerating is -10 m/s (negative sign indicates direction).
b) To calculate the distance John travels during the 5-second period of acceleration, we can use another kinematic equation:
\[s = v_i t + \frac{1}{2} a t^2\]
Where:
- \(s\) is the distance traveled
- \(v_i\) is the initial velocity (-10 m/s)
- \(a\) is the acceleration (2 m/s²)
- \(t\) is the time (5 seconds)
Plugging in the values:
\[s = (-10 \, \text{m/s}) \cdot (5 \, \text{seconds}) + \frac{1}{2} (2 \, \text{m/s}^2) \cdot (5 \, \text{seconds})^2\]
\[s = -50 \, \text{m} + 25 \, \text{m}\]
\[s = -25 \, \text{m}\]
So, during the 5-second period of acceleration, John travels a distance of 25 meters along the curved path in the opposite direction of his initial velocity.