The Problem:
A cyclist travels from A to B, a distance of 240 metres. He passes A at 12 m s1, maintains this speed for as long as he can, and then brakes so that he comes to a stop at B. If the maximum deceleration he can achieve when braking is 3 m s2, what is the least time in which he can get from A to B?
Second Method:
Cyclist Traveling from A to B
Cyclist Traveling from A to B
Given Data:
- Initial speed at A: 12 m/s
- Total distance to B: 240 meters
- Maximum deceleration: 3 m/s²
- We need to find the minimum time to reach B.
Step 1: Find the Minimum Braking Distance
Using the equation:
v² = u² + 2as
Since the cyclist stops at B, v = 0, u = 12 m/s, and a = -3 m/s²:
0 = 12² + 2(-3)s
0 = 144 – 6s
Solving for s:
s = 24 meters
Step 2: Distance Traveled at Full Speed
Total distance = 240 meters, braking distance = 24 meters.
Distance at full speed = 240 – 24 = 216 meters
Step 3: Find Time for Each Phase
Time at Full Speed:
Using the equation:
s = ut
216 = 12t
Solving for t:
t = 18 seconds
Time While Braking:
Using the equation:
v = u + at
0 = 12 – 3t
Solving for t:
t = 4 seconds
Final Answer
Total time taken:
t = 18 + 4 = 22 seconds
The least time to travel from A to B is 22 seconds. ✅
Third Method:
Cyclist Problem
A cyclist travels from point A to point B, a distance of 240 meters. He passes point A at 12 m/s, maintains this speed for as long as possible, and then brakes so that he comes to a stop at point B. If the maximum deceleration he can achieve when braking is 3 m/s², what is the least time in which he can get from A to B?
Solution
The least time in which the cyclist can travel from point A to point B is 22 seconds. This is calculated as follows:
- Phase 1 (Constant Speed): The cyclist travels at 12 m/s for 18 seconds, covering 216 meters.
- Phase 2 (Deceleration): The cyclist decelerates at 3 m/s² for 4 seconds, covering 24 meters.
Total time = 18 seconds + 4 seconds = 22 seconds.
Third Method:
Cyclist Traveling from A to B
Cyclist Traveling from A to B
Given Data:
- Initial speed at A: 12 m/s
- Total distance to B: 240 meters
- Maximum deceleration: 3 m/s²
- We need to find the minimum time to reach B.
Step 1: Find the Minimum Braking Distance
Using the equation:
v² = u² + 2as
Since the cyclist stops at B, v = 0, u = 12 m/s, and a = -3 m/s²:
0 = 12² + 2(-3)s
0 = 144 – 6s
Solving for s:
s = 24 meters
Step 2: Distance Traveled at Full Speed
Total distance = 240 meters, braking distance = 24 meters.
Distance at full speed = 240 – 24 = 216 meters
Step 3: Find Time for Each Phase
Time at Full Speed:
Using the equation:
s = ut
216 = 12t
Solving for t:
t = 18 seconds
Time While Braking:
Using the equation:
v = u + at
0 = 12 – 3t
Solving for t:
t = 4 seconds
Final Answer
Total time taken:
t = 18 + 4 = 22 seconds
The least time to travel from A to B is 22 seconds. ✅
