The Problem:
As a car passes the point A on a straight road, its speed is 10 m s. The car moves with constant acceleration am s2 along the road for T seconds until it reaches the point B, where its speed is V m s1. The car travels at this speed for a further 10 seconds, when it reaches the point C. From C it travels for a further T seconds with constant acceleration 3a ms2 until it reaches a speed of 20 m s1 at the point D. Sketch the (r,v) graph for the motion, and show that V = 12.5. Given that the distance between A and D is 675 m, find the values of a and T. (OCR)
Douglas Quadling Mechanics1 Miscellaneous Exercise1 Q4
As a car passes the point A on a straight road, its speed is 10 m s……..
Motion of a Car: Finding Acceleration, Time, and Speed
Introduction
In this problem, we analyze the motion of a car traveling along a straight road. The car undergoes different phases of motion, including acceleration, constant velocity, and further acceleration. Our goal is to determine the values of acceleration a and time T, and to verify that the speed V = 12.5 m/s.
Given Data:
- Initial speed at A: 10 m/s
- Acceleration from A to B: a m/s²
- Time taken from A to B: T seconds
- Speed at B: V m/s
- Constant speed phase from B to C: V m/s for 10 seconds
- Acceleration from C to D: 3a m/s²
- Time taken from C to D: T seconds
- Speed at D: 20 m/s
- Total distance: AD = 675 m
Step 1: Finding V
Using the first equation of motion:
v = u + at
For motion from A to B:
V = 10 + aT
For motion from C to D:
20 = V + (3a)T
Substituting V = 10 + aT into the second equation:
20 = (10 + aT) + 3aT
20 = 10 + aT + 3aT
10 = 4aT
aT = 2.5
Substituting aT = 2.5 into V = 10 + aT:
V = 10 + 2.5 = 12.5
Thus, we have verified that V = 12.5 m/s.
Step 2: Finding the Total Distance
The total distance AD = 675 m consists of three segments:
- Distance from A to B:
s₁ - Distance from B to C:
s₂ - Distance from C to D:
s₃
Distance from A to B:
Using the second equation of motion:
s₁ = uT + (1/2) aT²
s₁ = (10)T + (1/2) aT²
Distance from B to C:
s₂ = V × 10
s₂ = (12.5) × 10 = 125 m
Distance from C to D:
Using the equation:
s₃ = VT + (1/2) (3a) T²
s₃ = (12.5)T + (1/2) (3a) T²
Summing up all distances:
(10T + 0.5 aT²) + 125 + (12.5T + 1.5 aT²) = 675
10T + 0.5aT² + 125 + 12.5T + 1.5aT² = 675
22.5T + 2aT² = 550
Step 3: Solving for a and T
Using aT = 2.5, we substitute a = 2.5/T into:
22.5T + 2(2.5/T)T² = 550
22.5T + 5T = 550
27.5T = 550
T = 20 seconds
Now, using aT = 2.5:
a(20) = 2.5
a = 0.125 m/s²
Final Answers:
- Acceleration, a = 0.125 m/s²
- Time, T = 20 seconds
- Speed at B, V = 12.5 m/s (verified)
Velocity-Time Graph Sketch
The graph for this motion consists of three segments:
- A to B: Linearly increasing velocity from 10 m/s to 12.5 m/s.
- B to C: Constant velocity at 12.5 m/s.
- C to D: Linearly increasing velocity from 12.5 m/s to 20 m/s at a higher acceleration.
Conclusion
This problem demonstrates the importance of kinematic equations in analyzing motion with different acceleration phases. We successfully found the acceleration a = 0.125 m/s² and time T = 20 s, verifying that V = 12.5 m/s. The step-by-step breakdown ensures a clear understanding of the motion principles involved.
