Chasing a Starting Car Three Ways
This animation works through a classic pursuit problem: a car accelerates from rest at 1 m/s^2 while a person 60 m behind runs at constant speed, needing to stay within 20 m for at least 2 s to shout a warning. It compares three solution methods, algebraic equations, a velocity-time graph, and a critical-distance argument, to find the minimum chasing speed and the closest approach distance when running at 10 m/s. Useful for high school physics students studying kinematics and problem-solving strategies.
16:9 · Preview before teaching · automatic layout checks do not establish subject accuracy
The prompt that made it
（三种方法）汽车以 1 m/s 2的加速度启动，同时车后 60 m 远处有一人以一定的速度 v0匀速追赶要车停 下.已知在人离车的距离小于 20 m、持续时间为 2 s 喊停车才能把信息传给司机.问： （1）v0至少要多大？ （2）如果以 v0=10 m/s 的速度追车，人车距离最小为多少？
Make the next one in this series
- Show how the answer changes if the car's acceleration is doubled to 2 m/s^2
- Animate the case where the person starts running before the car begins moving
- Compare pursuit problems using relative velocity versus absolute distance-time graphs
